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1
docs/chapter_backtracking/n_queens_problem.md
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@ -269,6 +269,7 @@ comments: true
} }
*res = append(*res, newState) *res = append(*res, newState)
return
} }
// 遍历所有列 // 遍历所有列
for col := 0; col < n; col++ { for col := 0; col < n; col++ {

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docs/chapter_sorting/bubble_sort.md
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@ -276,7 +276,20 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="bubble_sort.rb" ```ruby title="bubble_sort.rb"
[class]{}-[func]{bubble_sort} ### 冒泡排序 ###
def bubble_sort(nums)
n = nums.length
# 外循环：未排序区间为 [0, i]
for i in (n - 1).downto(1)
# 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
for j in 0...i
if nums[j] > nums[j + 1]
# 交换 nums[j] 与 nums[j + 1]
nums[j], nums[j + 1] = nums[j + 1], nums[j]
end
end
end
end
``` ```
=== "Zig" === "Zig"
@ -587,7 +600,25 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="bubble_sort.rb" ```ruby title="bubble_sort.rb"
[class]{}-[func]{bubble_sort_with_flag} ### 冒泡排序（标志优化）###
def bubble_sort_with_flag(nums)
n = nums.length
# 外循环：未排序区间为 [0, i]
for i in (n - 1).downto(1)
flag = false # 初始化标志位
# 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
for j in 0...i
if nums[j] > nums[j + 1]
# 交换 nums[j] 与 nums[j + 1]
nums[j], nums[j + 1] = nums[j + 1], nums[j]
flag = true # 记录交换元素
end
end
break unless flag # 此轮“冒泡”未交换任何元素，直接跳出
end
end
``` ```
=== "Zig" === "Zig"

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docs/chapter_sorting/counting_sort.md
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@ -348,7 +348,25 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="counting_sort.rb" ```ruby title="counting_sort.rb"
[class]{}-[func]{counting_sort_naive} ### 计数排序 ###
def counting_sort_naive(nums)
# 简单实现，无法用于排序对象
# 1. 统计数组最大元素 m
m = 0
nums.each { |num| m = [m, num].max }
# 2. 统计各数字的出现次数
# counter[num] 代表 num 的出现次数
counter = Array.new(m + 1, 0)
nums.each { |num| counter[num] += 1 }
# 3. 遍历 counter ，将各元素填入原数组 nums
i = 0
for num in 0...(m + 1)
(0...counter[num]).each do
nums[i] = num
i += 1
end
end
end
``` ```
=== "Zig" === "Zig"
@ -854,7 +872,30 @@ $$
=== "Ruby" === "Ruby"
```ruby title="counting_sort.rb" ```ruby title="counting_sort.rb"
[class]{}-[func]{counting_sort} ### 计数排序 ###
def counting_sort(nums)
# 完整实现，可排序对象，并且是稳定排序
# 1. 统计数组最大元素 m
m = nums.max
# 2. 统计各数字的出现次数
# counter[num] 代表 num 的出现次数
counter = Array.new(m + 1, 0)
nums.each { |num| counter[num] += 1 }
# 3. 求 counter 的前缀和，将“出现次数”转换为“尾索引”
# 即 counter[num]-1 是 num 在 res 中最后一次出现的索引
(0...m).each { |i| counter[i + 1] += counter[i] }
# 4. 倒序遍历 nums, 将各元素填入结果数组 res
# 初始化数组 res 用于记录结果
n = nums.length
res = Array.new(n, 0)
(n - 1).downto(0).each do |i|
num = nums[i]
res[counter[num] - 1] = num # 将 num 放置到对应索引处
counter[num] -= 1 # 令前缀和自减 1 ，得到下次放置 num 的索引
end
# 使用结果数组 res 覆盖原数组 nums
(0...n).each { |i| nums[i] = res[i] }
end
``` ```
=== "Zig" === "Zig"

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docs/chapter_sorting/quick_sort.md
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@ -353,7 +353,24 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSort}-[func]{partition} ### 哨兵划分 ###
def partition(nums, left, right)
# 以 nums[left] 为基准数
i, j = left, right
while i < j
while i < j && nums[j] >= nums[left]
j -= 1 # 从右向左找首个小于基准数的元素
end
while i < j && nums[i] <= nums[left]
i += 1 # 从左向右找首个大于基准数的元素
end
# 元素交换
nums[i], nums[j] = nums[j], nums[i]
end
# 将基准数交换至两子数组的分界线
nums[i], nums[left] = nums[left], nums[i]
i # 返回基准数的索引
end
``` ```
=== "Zig" === "Zig"
@ -594,7 +611,18 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSort}-[func]{quick_sort} ### 快速排序类 ###
def quick_sort(nums, left, right)
# 子数组长度不为 1 时递归
if left < right
# 哨兵划分
pivot = partition(nums, left, right)
# 递归左子数组、右子数组
quick_sort(nums, left, pivot - 1)
quick_sort(nums, pivot + 1, right)
end
nums
end
``` ```
=== "Zig" === "Zig"
@ -1067,9 +1095,38 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSortMedian}-[func]{median_three} ### 选取三个候选元素的中位数 ###
def median_three(nums, left, mid, right)
[class]{QuickSortMedian}-[func]{partition} # 选取三个候选元素的中位数
_l, _m, _r = nums[left], nums[mid], nums[right]
# m 在 l 和 r 之间
return mid if (_l <= _m && _m <= _r) || (_r <= _m && _m <= _l)
# l 在 m 和 r 之间
return left if (_m <= _l && _l <= _r) || (_r <= _l && _l <= _m)
return right
end
### 哨兵划分（三数取中值）###
def partition(nums, left, right)
### 以 nums[left] 为基准数
med = median_three(nums, left, (left + right) / 2, right)
# 将中位数交换至数组最左断
nums[left], nums[med] = nums[med], nums[left]
i, j = left, right
while i < j
while i < j && nums[j] >= nums[left]
j -= 1 # 从右向左找首个小于基准数的元素
end
while i < j && nums[i] <= nums[left]
i += 1 # 从左向右找首个大于基准数的元素
end
# 元素交换
nums[i], nums[j] = nums[j], nums[i]
end
# 将基准数交换至两子数组的分界线
nums[i], nums[left] = nums[left], nums[i]
i # 返回基准数的索引
end
``` ```
=== "Zig" === "Zig"
@ -1377,7 +1434,22 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSortTailCall}-[func]{quick_sort} ### 快速排序（尾递归优化）###
def quick_sort(nums, left, right)
# 子数组长度不为 1 时递归
while left < right
# 哨兵划分
pivot = partition(nums, left, right)
# 对两个子数组中较短的那个执行快速排序
if pivot - left < right - pivot
quick_sort(nums, left, pivot - 1)
left = pivot + 1 # 剩余未排序区间为 [pivot + 1, right]
else
quick_sort(nums, pivot + 1, right)
right = pivot - 1 # 剩余未排序区间为 [left, pivot - 1]
end
end
end
``` ```
=== "Zig" === "Zig"

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docs/chapter_sorting/radix_sort.md
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@ -677,11 +677,51 @@ $$
=== "Ruby" === "Ruby"
```ruby title="radix_sort.rb" ```ruby title="radix_sort.rb"
[class]{}-[func]{digit} ### 获取元素 num 的第 k 位，其中 exp = 10^(k-1) ###
def digit(num, exp)
[class]{}-[func]{counting_sort_digit} # 转入 exp 而非 k 可以避免在此重复执行昂贵的次方计算
(num / exp) % 10
[class]{}-[func]{radix_sort} end
### 计数排序（根据 nums 第 k 位排序）###
def counting_sort_digit(nums, exp)
# 十进制的位范围为 0~9 ，因此需要长度为 10 的桶数组
counter = Array.new(10, 0)
n = nums.length
# 统计 0~9 各数字的出现次数
for i in 0...n
d = digit(nums[i], exp) # 获取 nums[i] 第 k 位，记为 d
counter[d] += 1 # 统计数字 d 的出现次数
end
# 求前缀和，将“出现个数”转换为“数组索引”
(1...10).each { |i| counter[i] += counter[i - 1] }
# 倒序遍历，根据桶内统计结果，将各元素填入 res
res = Array.new(n, 0)
for i in (n - 1).downto(0)
d = digit(nums[i], exp)
j = counter[d] - 1 # 获取 d 在数组中的索引 j
res[j] = nums[i] # 将当前元素填入索引 j
counter[d] -= 1 # 将 d 的数量减 1
end
# 使用结果覆盖原数组 nums
(0...n).each { |i| nums[i] = res[i] }
end
### 基数排序 ###
def radix_sort(nums)
# 获取数组的最大元素，用于判断最大位数
m = nums.max
# 按照从低位到高位的顺序遍历
exp = 1
while exp <= m
# 对数组元素的第 k 位执行计数排序
# k = 1 -> exp = 1
# k = 2 -> exp = 10
# 即 exp = 10^(k-1)
counting_sort_digit(nums, exp)
exp *= 10
end
end
``` ```
=== "Zig" === "Zig"

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docs/chapter_sorting/selection_sort.md
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@ -306,7 +306,22 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="selection_sort.rb" ```ruby title="selection_sort.rb"
[class]{}-[func]{selection_sort} ### 选择排序 ###
def selection_sort(nums)
n = nums.length
# 外循环：未排序区间为 [i, n-1]
for i in 0...(n - 1)
# 内循环：找到未排序区间内的最小元素
k = i
for j in (i + 1)...n
if nums[j] < nums[k]
k = j # 记录最小元素的索引
end
end
# 将该最小元素与未排序区间的首个元素交换
nums[i], nums[k] = nums[k], nums[i]
end
end
``` ```
=== "Zig" === "Zig"

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en/docs/chapter_backtracking/n_queens_problem.md
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@ -269,6 +269,7 @@ Please note, in an $n$-dimensional matrix, the range of $row - col$ is $[-n + 1,
} }
*res = append(*res, newState) *res = append(*res, newState)
return
} }
// 遍历所有列 // 遍历所有列
for col := 0; col < n; col++ { for col := 0; col < n; col++ {

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en/docs/chapter_computational_complexity/space_complexity.md
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@ -735,8 +735,8 @@ Let the size of the input data be $n$, the following chart displays common types
$$ $$
\begin{aligned} \begin{aligned}
O(1) < O(\log n) < O(n) < O(n^2) < O(2^n) \newline & O(1) < O(\log n) < O(n) < O(n^2) < O(2^n) \newline
\text{Constant Order} < \text{Logarithmic Order} < \text{Linear Order} < \text{Quadratic Order} < \text{Exponential Order} & \text{Constant} < \text{Logarithmic} < \text{Linear} < \text{Quadratic} < \text{Exponential}
\end{aligned} \end{aligned}
$$ $$

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en/docs/chapter_computational_complexity/time_complexity.md
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@ -967,8 +967,8 @@ Let's consider the input data size as $n$. The common types of time complexities
$$ $$
\begin{aligned} \begin{aligned}
O(1) < O(\log n) < O(n) < O(n \log n) < O(n^2) < O(2^n) < O(n!) \newline & O(1) < O(\log n) < O(n) < O(n \log n) < O(n^2) < O(2^n) < O(n!) \newline
\text{Constant Order} < \text{Logarithmic Order} < \text{Linear Order} < \text{Linear-Logarithmic Order} < \text{Quadratic Order} < \text{Exponential Order} < \text{Factorial Order} & \text{Constant} < \text{Log} < \text{Linear} < \text{Linear-Log} < \text{Quadratic} < \text{Exp} < \text{Factorial}
\end{aligned} \end{aligned}
$$ $$

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en/docs/chapter_graph/graph_traversal.md
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@ -24,7 +24,7 @@ BFS is usually implemented with the help of a queue, as shown in the code below.
2. In each iteration of the loop, pop the vertex at the front of the queue and record it as visited, then add all adjacent vertices of that vertex to the back of the queue. 2. In each iteration of the loop, pop the vertex at the front of the queue and record it as visited, then add all adjacent vertices of that vertex to the back of the queue.
3. Repeat step `2.` until all vertices have been visited. 3. Repeat step `2.` until all vertices have been visited.
To prevent revisiting vertices, we use a hash table `visited` to record which nodes have been visited. To prevent revisiting vertices, we use a hash set `visited` to record which nodes have been visited.
=== "Python" === "Python"
@ -528,7 +528,7 @@ The code is relatively abstract, it is suggested to compare with Figure 9-10 to
**Time complexity**: All vertices will be enqueued and dequeued once, using $O(|V|)$ time; in the process of traversing adjacent vertices, since it is an undirected graph, all edges will be visited $2$ times, using $O(2|E|)$ time; overall using $O(|V| + |E|)$ time. **Time complexity**: All vertices will be enqueued and dequeued once, using $O(|V|)$ time; in the process of traversing adjacent vertices, since it is an undirected graph, all edges will be visited $2$ times, using $O(2|E|)$ time; overall using $O(|V| + |E|)$ time.
**Space complexity**: The maximum number of vertices in list `res`, hash table `visited`, and queue `que` is $|V|$, using $O(|V|)$ space. **Space complexity**: The maximum number of vertices in list `res`, hash set `visited`, and queue `que` is $|V|$, using $O(|V|)$ space.
## 9.3.2 &nbsp; Depth-first search ## 9.3.2 &nbsp; Depth-first search
@ -540,7 +540,7 @@ The code is relatively abstract, it is suggested to compare with Figure 9-10 to
### 1. &nbsp; Algorithm implementation ### 1. &nbsp; Algorithm implementation
This "go as far as possible and then return" algorithm paradigm is usually implemented based on recursion. Similar to breadth-first search, in depth-first search, we also need the help of a hash table `visited` to record the visited vertices to avoid revisiting. This "go as far as possible and then return" algorithm paradigm is usually implemented based on recursion. Similar to breadth-first search, in depth-first search, we also need the help of a hash set `visited` to record the visited vertices to avoid revisiting.
=== "Python" === "Python"
@ -1003,4 +1003,4 @@ To deepen the understanding, it is suggested to combine Figure 9-12 with the cod
**Time complexity**: All vertices will be visited once, using $O(|V|)$ time; all edges will be visited twice, using $O(2|E|)$ time; overall using $O(|V| + |E|)$ time. **Time complexity**: All vertices will be visited once, using $O(|V|)$ time; all edges will be visited twice, using $O(2|E|)$ time; overall using $O(|V| + |E|)$ time.
**Space complexity**: The maximum number of vertices in list `res`, hash table `visited` is $|V|$, and the maximum recursion depth is $|V|$, therefore using $O(|V|)$ space. **Space complexity**: The maximum number of vertices in list `res`, hash set `visited` is $|V|$, and the maximum recursion depth is $|V|$, therefore using $O(|V|)$ space.

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en/docs/chapter_hashing/hash_collision.md
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35
en/docs/chapter_sorting/bubble_sort.md
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@ -276,7 +276,20 @@ Example code is as follows:
=== "Ruby" === "Ruby"
```ruby title="bubble_sort.rb" ```ruby title="bubble_sort.rb"
[class]{}-[func]{bubble_sort} ### 冒泡排序 ###
def bubble_sort(nums)
n = nums.length
# 外循环：未排序区间为 [0, i]
for i in (n - 1).downto(1)
# 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
for j in 0...i
if nums[j] > nums[j + 1]
# 交换 nums[j] 与 nums[j + 1]
nums[j], nums[j + 1] = nums[j + 1], nums[j]
end
end
end
end
``` ```
=== "Zig" === "Zig"
@ -587,7 +600,25 @@ Even after optimization, the worst-case time complexity and average time complex
=== "Ruby" === "Ruby"
```ruby title="bubble_sort.rb" ```ruby title="bubble_sort.rb"
[class]{}-[func]{bubble_sort_with_flag} ### 冒泡排序（标志优化）###
def bubble_sort_with_flag(nums)
n = nums.length
# 外循环：未排序区间为 [0, i]
for i in (n - 1).downto(1)
flag = false # 初始化标志位
# 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
for j in 0...i
if nums[j] > nums[j + 1]
# 交换 nums[j] 与 nums[j + 1]
nums[j], nums[j + 1] = nums[j + 1], nums[j]
flag = true # 记录交换元素
end
end
break unless flag # 此轮“冒泡”未交换任何元素，直接跳出
end
end
``` ```
=== "Zig" === "Zig"

45
en/docs/chapter_sorting/counting_sort.md
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@ -348,7 +348,25 @@ The code is shown below:
=== "Ruby" === "Ruby"
```ruby title="counting_sort.rb" ```ruby title="counting_sort.rb"
[class]{}-[func]{counting_sort_naive} ### 计数排序 ###
def counting_sort_naive(nums)
# 简单实现，无法用于排序对象
# 1. 统计数组最大元素 m
m = 0
nums.each { |num| m = [m, num].max }
# 2. 统计各数字的出现次数
# counter[num] 代表 num 的出现次数
counter = Array.new(m + 1, 0)
nums.each { |num| counter[num] += 1 }
# 3. 遍历 counter ，将各元素填入原数组 nums
i = 0
for num in 0...(m + 1)
(0...counter[num]).each do
nums[i] = num
i += 1
end
end
end
``` ```
=== "Zig" === "Zig"
@ -854,7 +872,30 @@ The implementation code of counting sort is shown below:
=== "Ruby" === "Ruby"
```ruby title="counting_sort.rb" ```ruby title="counting_sort.rb"
[class]{}-[func]{counting_sort} ### 计数排序 ###
def counting_sort(nums)
# 完整实现，可排序对象，并且是稳定排序
# 1. 统计数组最大元素 m
m = nums.max
# 2. 统计各数字的出现次数
# counter[num] 代表 num 的出现次数
counter = Array.new(m + 1, 0)
nums.each { |num| counter[num] += 1 }
# 3. 求 counter 的前缀和，将“出现次数”转换为“尾索引”
# 即 counter[num]-1 是 num 在 res 中最后一次出现的索引
(0...m).each { |i| counter[i + 1] += counter[i] }
# 4. 倒序遍历 nums, 将各元素填入结果数组 res
# 初始化数组 res 用于记录结果
n = nums.length
res = Array.new(n, 0)
(n - 1).downto(0).each do |i|
num = nums[i]
res[counter[num] - 1] = num # 将 num 放置到对应索引处
counter[num] -= 1 # 令前缀和自减 1 ，得到下次放置 num 的索引
end
# 使用结果数组 res 覆盖原数组 nums
(0...n).each { |i| nums[i] = res[i] }
end
``` ```
=== "Zig" === "Zig"

84
en/docs/chapter_sorting/quick_sort.md
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@ -353,7 +353,24 @@ After the pivot partitioning, the original array is divided into three parts: le
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSort}-[func]{partition} ### 哨兵划分 ###
def partition(nums, left, right)
# 以 nums[left] 为基准数
i, j = left, right
while i < j
while i < j && nums[j] >= nums[left]
j -= 1 # 从右向左找首个小于基准数的元素
end
while i < j && nums[i] <= nums[left]
i += 1 # 从左向右找首个大于基准数的元素
end
# 元素交换
nums[i], nums[j] = nums[j], nums[i]
end
# 将基准数交换至两子数组的分界线
nums[i], nums[left] = nums[left], nums[i]
i # 返回基准数的索引
end
``` ```
=== "Zig" === "Zig"
@ -594,7 +611,18 @@ The overall process of quick sort is shown in Figure 11-9.
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSort}-[func]{quick_sort} ### 快速排序类 ###
def quick_sort(nums, left, right)
# 子数组长度不为 1 时递归
if left < right
# 哨兵划分
pivot = partition(nums, left, right)
# 递归左子数组、右子数组
quick_sort(nums, left, pivot - 1)
quick_sort(nums, pivot + 1, right)
end
nums
end
``` ```
=== "Zig" === "Zig"
@ -1067,9 +1095,38 @@ Sample code is as follows:
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSortMedian}-[func]{median_three} ### 选取三个候选元素的中位数 ###
def median_three(nums, left, mid, right)
[class]{QuickSortMedian}-[func]{partition} # 选取三个候选元素的中位数
_l, _m, _r = nums[left], nums[mid], nums[right]
# m 在 l 和 r 之间
return mid if (_l <= _m && _m <= _r) || (_r <= _m && _m <= _l)
# l 在 m 和 r 之间
return left if (_m <= _l && _l <= _r) || (_r <= _l && _l <= _m)
return right
end
### 哨兵划分（三数取中值）###
def partition(nums, left, right)
### 以 nums[left] 为基准数
med = median_three(nums, left, (left + right) / 2, right)
# 将中位数交换至数组最左断
nums[left], nums[med] = nums[med], nums[left]
i, j = left, right
while i < j
while i < j && nums[j] >= nums[left]
j -= 1 # 从右向左找首个小于基准数的元素
end
while i < j && nums[i] <= nums[left]
i += 1 # 从左向右找首个大于基准数的元素
end
# 元素交换
nums[i], nums[j] = nums[j], nums[i]
end
# 将基准数交换至两子数组的分界线
nums[i], nums[left] = nums[left], nums[i]
i # 返回基准数的索引
end
``` ```
=== "Zig" === "Zig"
@ -1377,7 +1434,22 @@ To prevent the accumulation of stack frame space, we can compare the lengths of
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSortTailCall}-[func]{quick_sort} ### 快速排序（尾递归优化）###
def quick_sort(nums, left, right)
# 子数组长度不为 1 时递归
while left < right
# 哨兵划分
pivot = partition(nums, left, right)
# 对两个子数组中较短的那个执行快速排序
if pivot - left < right - pivot
quick_sort(nums, left, pivot - 1)
left = pivot + 1 # 剩余未排序区间为 [pivot + 1, right]
else
quick_sort(nums, pivot + 1, right)
right = pivot - 1 # 剩余未排序区间为 [left, pivot - 1]
end
end
end
``` ```
=== "Zig" === "Zig"

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en/docs/chapter_sorting/radix_sort.md
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@ -677,11 +677,51 @@ Additionally, we need to slightly modify the counting sort code to allow sorting
=== "Ruby" === "Ruby"
```ruby title="radix_sort.rb" ```ruby title="radix_sort.rb"
[class]{}-[func]{digit} ### 获取元素 num 的第 k 位，其中 exp = 10^(k-1) ###
def digit(num, exp)
[class]{}-[func]{counting_sort_digit} # 转入 exp 而非 k 可以避免在此重复执行昂贵的次方计算
(num / exp) % 10
[class]{}-[func]{radix_sort} end
### 计数排序（根据 nums 第 k 位排序）###
def counting_sort_digit(nums, exp)
# 十进制的位范围为 0~9 ，因此需要长度为 10 的桶数组
counter = Array.new(10, 0)
n = nums.length
# 统计 0~9 各数字的出现次数
for i in 0...n
d = digit(nums[i], exp) # 获取 nums[i] 第 k 位，记为 d
counter[d] += 1 # 统计数字 d 的出现次数
end
# 求前缀和，将“出现个数”转换为“数组索引”
(1...10).each { |i| counter[i] += counter[i - 1] }
# 倒序遍历，根据桶内统计结果，将各元素填入 res
res = Array.new(n, 0)
for i in (n - 1).downto(0)
d = digit(nums[i], exp)
j = counter[d] - 1 # 获取 d 在数组中的索引 j
res[j] = nums[i] # 将当前元素填入索引 j
counter[d] -= 1 # 将 d 的数量减 1
end
# 使用结果覆盖原数组 nums
(0...n).each { |i| nums[i] = res[i] }
end
### 基数排序 ###
def radix_sort(nums)
# 获取数组的最大元素，用于判断最大位数
m = nums.max
# 按照从低位到高位的顺序遍历
exp = 1
while exp <= m
# 对数组元素的第 k 位执行计数排序
# k = 1 -> exp = 1
# k = 2 -> exp = 10
# 即 exp = 10^(k-1)
counting_sort_digit(nums, exp)
exp *= 10
end
end
``` ```
=== "Zig" === "Zig"

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en/docs/chapter_sorting/selection_sort.md
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@ -306,7 +306,22 @@ In the code, we use $k$ to record the smallest element within the unsorted inter
=== "Ruby" === "Ruby"
```ruby title="selection_sort.rb" ```ruby title="selection_sort.rb"
[class]{}-[func]{selection_sort} ### 选择排序 ###
def selection_sort(nums)
n = nums.length
# 外循环：未排序区间为 [i, n-1]
for i in 0...(n - 1)
# 内循环：找到未排序区间内的最小元素
k = i
for j in (i + 1)...n
if nums[j] < nums[k]
k = j # 记录最小元素的索引
end
end
# 将该最小元素与未排序区间的首个元素交换
nums[i], nums[k] = nums[k], nums[i]
end
end
``` ```
=== "Zig" === "Zig"

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en/docs/chapter_tree/binary_tree.md
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@ -4,7 +4,7 @@ comments: true
# 7.1 &nbsp; Binary tree # 7.1 &nbsp; Binary tree
A <u>binary tree</u> is a non-linear data structure that represents the ancestral and descendent relationships, embodying the "divide and conquer" logic. Similar to a linked list, the basic unit of a binary tree is a node, each containing a value, a reference to the left child node, and a reference to the right child node. A <u>binary tree</u> is a non-linear data structure that represents the hierarchical relationship between ancestors and descendants, embodying the divide-and-conquer logic of "splitting into two". Similar to a linked list, the basic unit of a binary tree is a node, each containing a value, a reference to the left child node, and a reference to the right child node.
=== "Python" === "Python"
@ -218,7 +218,7 @@ The commonly used terminology of binary trees is shown in Figure 7-2.
- <u>Leaf node</u>: A node with no children, both of its pointers point to `None`. - <u>Leaf node</u>: A node with no children, both of its pointers point to `None`.
- <u>Edge</u>: The line segment connecting two nodes, i.e., node reference (pointer). - <u>Edge</u>: The line segment connecting two nodes, i.e., node reference (pointer).
- The <u>level</u> of a node: Incrementing from top to bottom, with the root node's level being 1. - The <u>level</u> of a node: Incrementing from top to bottom, with the root node's level being 1.
- The <u>degree</u> of a node: The number of a node's children. In a binary tree, the degree can be 0, 1, or 2. - The <u>degree</u> of a node: The number of children a node has. In a binary tree, the degree can be 0, 1, or 2.
- The <u>height</u> of a binary tree: The number of edges passed from the root node to the farthest leaf node. - The <u>height</u> of a binary tree: The number of edges passed from the root node to the farthest leaf node.
- The <u>depth</u> of a node: The number of edges passed from the root node to the node. - The <u>depth</u> of a node: The number of edges passed from the root node to the node.
- The <u>height</u> of a node: The number of edges from the farthest leaf node to the node. - The <u>height</u> of a node: The number of edges from the farthest leaf node to the node.
@ -229,13 +229,13 @@ The commonly used terminology of binary trees is shown in Figure 7-2.
!!! tip !!! tip
Please note that we usually define "height" and "depth" as "the number of edges passed," but some problems or textbooks may define them as "the number of nodes passed." In this case, both height and depth need to be incremented by 1. Please note that we typically define "height" and "depth" as "the number of edges traversed", but some problems or textbooks may define them as "the number of nodes traversed". In such cases, both height and depth need to be incremented by 1.
## 7.1.2 &nbsp; Basic operations of binary trees ## 7.1.2 &nbsp; Basic operations of binary trees
### 1. &nbsp; Initializing a binary tree ### 1. &nbsp; Initializing a binary tree
Similar to a linked list, initialize nodes first, then construct references (pointers). Similar to a linked list, begin by initialize nodes, then construct references (pointers).
=== "Python" === "Python"
@ -619,13 +619,13 @@ Similar to a linked list, inserting and removing nodes in a binary tree can be a
!!! tip !!! tip
It's important to note that inserting nodes may change the original logical structure of the binary tree, while removing nodes usually means removing the node and all its subtrees. Therefore, in a binary tree, insertion and removal are usually performed through a set of operations to achieve meaningful actions. It's important to note that inserting nodes may change the original logical structure of the binary tree, while removing nodes typically involves removing the node and all its subtrees. Therefore, in a binary tree, insertion and removal are usually performed through a coordinated set of operations to achieve meaningful outcomes.
## 7.1.3 &nbsp; Common types of binary trees ## 7.1.3 &nbsp; Common types of binary trees
### 1. &nbsp; Perfect binary tree ### 1. &nbsp; Perfect binary tree
As shown in Figure 7-4, in a <u>perfect binary tree</u>, all levels of nodes are fully filled. In a perfect binary tree, the degree of leaf nodes is $0$, and the degree of all other nodes is $2$; if the tree's height is $h$, then the total number of nodes is $2^{h+1} - 1$, showing a standard exponential relationship, reflecting the common phenomenon of cell division in nature. As shown in Figure 7-4, in a <u>perfect binary tree</u>, all levels of nodes are fully filled. In a perfect binary tree, the degree of leaf nodes is $0$, while the degree of all other nodes is $2$; if the tree's height is $h$, then the total number of nodes is $2^{h+1} - 1$, showing a standard exponential relationship, reflecting the common phenomenon of cell division in nature.
!!! tip !!! tip
@ -661,7 +661,7 @@ As shown in Figure 7-7, in a <u>balanced binary tree</u>, the absolute differenc
## 7.1.4 &nbsp; Degeneration of binary trees ## 7.1.4 &nbsp; Degeneration of binary trees
Figure 7-8 shows the ideal and degenerate structures of binary trees. When every level of a binary tree is filled, it reaches the "perfect binary tree"; when all nodes are biased towards one side, the binary tree degenerates into a "linked list". Figure 7-8 shows the ideal and degenerate structures of binary trees. A binary tree becomes a "perfect binary tree" when every level is filled; while it degenerates into a "linked list" when all nodes are biased toward one side.
- The perfect binary tree is the ideal situation, fully leveraging the "divide and conquer" advantage of binary trees. - The perfect binary tree is the ideal situation, fully leveraging the "divide and conquer" advantage of binary trees.
- A linked list is another extreme, where operations become linear, degrading the time complexity to $O(n)$. - A linked list is another extreme, where operations become linear, degrading the time complexity to $O(n)$.

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@ -26,7 +26,7 @@ comments: true
也就是說，我們可以採取逐行放置策略：從第一行開始，在每行放置一個皇后，直至最後一行結束。 也就是說，我們可以採取逐行放置策略：從第一行開始，在每行放置一個皇后，直至最後一行結束。
圖 13-17 所示為 $4$ 皇后問題的逐行放置過程。受畫幅限制，圖 13-17 僅展開了第一行的其中一個搜尋分支，並且將不滿足列約束和對角線約束的方案都進行了剪枝。 圖 13-17 所示為 4 皇后問題的逐行放置過程。受畫幅限制，圖 13-17 僅展開了第一行的其中一個搜尋分支，並且將不滿足列約束和對角線約束的方案都進行了剪枝。
![逐行放置策略](n_queens_problem.assets/n_queens_placing.png){ class="animation-figure" } ![逐行放置策略](n_queens_problem.assets/n_queens_placing.png){ class="animation-figure" }
@ -269,6 +269,7 @@ comments: true
} }
*res = append(*res, newState) *res = append(*res, newState)
return
} }
// 走訪所有列 // 走訪所有列
for col := 0; col < n; col++ { for col := 0; col < n; col++ {

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@ -724,7 +724,7 @@ comments: true
(*selected)[i] = true (*selected)[i] = true
*state = append(*state, choice) *state = append(*state, choice)
// 進行下一輪選擇 // 進行下一輪選擇
backtrackI(state, choices, selected, res) backtrackII(state, choices, selected, res)
// 回退：撤銷選擇，恢復到之前的狀態 // 回退：撤銷選擇，恢復到之前的狀態
(*selected)[i] = false (*selected)[i] = false
*state = (*state)[:len(*state)-1] *state = (*state)[:len(*state)-1]

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zh-Hant/docs/chapter_sorting/bubble_sort.md
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@ -276,7 +276,20 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="bubble_sort.rb" ```ruby title="bubble_sort.rb"
[class]{}-[func]{bubble_sort} ### 泡沫排序 ###
def bubble_sort(nums)
n = nums.length
# 外迴圈：未排序區間為 [0, i]
for i in (n - 1).downto(1)
# 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
for j in 0...i
if nums[j] > nums[j + 1]
# 交換 nums[j] 與 nums[j + 1]
nums[j], nums[j + 1] = nums[j + 1], nums[j]
end
end
end
end
``` ```
=== "Zig" === "Zig"
@ -587,7 +600,25 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="bubble_sort.rb" ```ruby title="bubble_sort.rb"
[class]{}-[func]{bubble_sort_with_flag} ### 泡沫排序（標誌最佳化）###
def bubble_sort_with_flag(nums)
n = nums.length
# 外迴圈：未排序區間為 [0, i]
for i in (n - 1).downto(1)
flag = false # 初始化標誌位
# 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
for j in 0...i
if nums[j] > nums[j + 1]
# 交換 nums[j] 與 nums[j + 1]
nums[j], nums[j + 1] = nums[j + 1], nums[j]
flag = true # 記錄交換元素
end
end
break unless flag # 此輪“冒泡”未交換任何元素，直接跳出
end
end
``` ```
=== "Zig" === "Zig"

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zh-Hant/docs/chapter_sorting/bucket_sort.md
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@ -406,7 +406,35 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="bucket_sort.rb" ```ruby title="bucket_sort.rb"
[class]{}-[func]{bucket_sort} ### 桶排序 ###
def bucket_sort(nums)
# 初始化 k = n/2 個桶，預期向每個桶分配 2 個元素
k = nums.length / 2
buckets = Array.new(k) { [] }
# 1. 將陣列元素分配到各個桶中
nums.each do |num|
# 輸入資料範圍為 [0, 1)，使用 num * k 對映到索引範圍 [0, k-1]
i = (num * k).to_i
# 將 num 新增進桶 i
buckets[i] << num
end
# 2. 對各個桶執行排序
buckets.each do |bucket|
# 使用內建排序函式，也可以替換成其他排序演算法
bucket.sort!
end
# 3. 走訪桶合併結果
i = 0
buckets.each do |bucket|
bucket.each do |num|
nums[i] = num
i += 1
end
end
end
``` ```
=== "Zig" === "Zig"

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zh-Hant/docs/chapter_sorting/counting_sort.md
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@ -348,7 +348,25 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="counting_sort.rb" ```ruby title="counting_sort.rb"
[class]{}-[func]{counting_sort_naive} ### 計數排序 ###
def counting_sort_naive(nums)
# 簡單實現，無法用於排序物件
# 1. 統計陣列最大元素 m
m = 0
nums.each { |num| m = [m, num].max }
# 2. 統計各數字的出現次數
# counter[num] 代表 num 的出現次數
counter = Array.new(m + 1, 0)
nums.each { |num| counter[num] += 1 }
# 3. 走訪 counter ，將各元素填入原陣列 nums
i = 0
for num in 0...(m + 1)
(0...counter[num]).each do
nums[i] = num
i += 1
end
end
end
``` ```
=== "Zig" === "Zig"
@ -854,7 +872,30 @@ $$
=== "Ruby" === "Ruby"
```ruby title="counting_sort.rb" ```ruby title="counting_sort.rb"
[class]{}-[func]{counting_sort} ### 計數排序 ###
def counting_sort(nums)
# 完整實現，可排序物件，並且是穩定排序
# 1. 統計陣列最大元素 m
m = nums.max
# 2. 統計各數字的出現次數
# counter[num] 代表 num 的出現次數
counter = Array.new(m + 1, 0)
nums.each { |num| counter[num] += 1 }
# 3. 求 counter 的前綴和，將“出現次數”轉換為“尾索引”
# 即 counter[num]-1 是 num 在 res 中最後一次出現的索引
(0...m).each { |i| counter[i + 1] += counter[i] }
# 4. 倒序走訪 nums, 將各元素填入結果陣列 res
# 初始化陣列 res 用於記錄結果
n = nums.length
res = Array.new(n, 0)
(n - 1).downto(0).each do |i|
num = nums[i]
res[counter[num] - 1] = num # 將 num 放置到對應索引處
counter[num] -= 1 # 令前綴和自減 1 ，得到下次放置 num 的索引
end
# 使用結果陣列 res 覆蓋原陣列 nums
(0...n).each { |i| nums[i] = res[i] }
end
``` ```
=== "Zig" === "Zig"

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zh-Hant/docs/chapter_sorting/heap_sort.md
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@ -582,9 +582,38 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="heap_sort.rb" ```ruby title="heap_sort.rb"
[class]{}-[func]{sift_down} ### 堆積的長度為 n ，從節點 i 開始，從頂至底堆積化 ###
def sift_down(nums, n, i)
[class]{}-[func]{heap_sort} while true
# 判斷節點 i, l, r 中值最大的節點，記為 ma
l = 2 * i + 1
r = 2 * i + 2
ma = i
ma = l if l < n && nums[l] > nums[ma]
ma = r if r < n && nums[r] > nums[ma]
# 若節點 i 最大或索引 l, r 越界，則無須繼續堆積化，跳出
break if ma == i
# 交換兩節點
nums[i], nums[ma] = nums[ma], nums[i]
# 迴圈向下堆積化
i = ma
end
end
### 堆積排序 ###
def heap_sort(nums)
# 建堆積操作：堆積化除葉節點以外的其他所有節點
(nums.length / 2 - 1).downto(0) do |i|
sift_down(nums, nums.length, i)
end
# 從堆積中提取最大元素，迴圈 n-1 輪
(nums.length - 1).downto(1) do |i|
# 交換根節點與最右葉節點（交換首元素與尾元素）
nums[0], nums[i] = nums[i], nums[0]
# 以根節點為起點，從頂至底進行堆積化
sift_down(nums, i, 0)
end
end
``` ```
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zh-Hant/docs/chapter_sorting/merge_sort.md
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@ -630,9 +630,53 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="merge_sort.rb" ```ruby title="merge_sort.rb"
[class]{}-[func]{merge} ### 合併左子陣列和右子陣列 ###
def merge(nums, left, mid, right)
[class]{}-[func]{merge_sort} # 左子陣列區間為 [left, mid], 右子陣列區間為 [mid+1, right]
# 建立一個臨時陣列 tmp，用於存放合併後的結果
tmp = Array.new(right - left + 1, 0)
# 初始化左子陣列和右子陣列的起始索引
i, j, k = left, mid + 1, 0
# 當左右子陣列都還有元素時，進行比較並將較小的元素複製到臨時陣列中
while i <= mid && j <= right
if nums[i] <= nums[j]
tmp[k] = nums[i]
i += 1
else
tmp[k] = nums[j]
j += 1
end
k += 1
end
# 將左子陣列和右子陣列的剩餘元素複製到臨時陣列中
while i <= mid
tmp[k] = nums[i]
i += 1
k += 1
end
while j <= right
tmp[k] = nums[j]
j += 1
k += 1
end
# 將臨時陣列 tmp 中的元素複製回原陣列 nums 的對應區間
(0...tmp.length).each do |k|
nums[left + k] = tmp[k]
end
end
### 合併排序 ###
def merge_sort(nums, left, right)
# 終止條件
# 當子陣列長度為 1 時終止遞迴
return if left >= right
# 劃分階段
mid = (left + right) / 2 # 計算中點
merge_sort(nums, left, mid) # 遞迴左子陣列
merge_sort(nums, mid + 1, right) # 遞迴右子陣列
# 合併階段
merge(nums, left, mid, right)
end
``` ```
=== "Zig" === "Zig"

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@ -353,7 +353,24 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSort}-[func]{partition} ### 哨兵劃分 ###
def partition(nums, left, right)
# 以 nums[left] 為基準數
i, j = left, right
while i < j
while i < j && nums[j] >= nums[left]
j -= 1 # 從右向左找首個小於基準數的元素
end
while i < j && nums[i] <= nums[left]
i += 1 # 從左向右找首個大於基準數的元素
end
# 元素交換
nums[i], nums[j] = nums[j], nums[i]
end
# 將基準數交換至兩子陣列的分界線
nums[i], nums[left] = nums[left], nums[i]
i # 返回基準數的索引
end
``` ```
=== "Zig" === "Zig"
@ -594,7 +611,18 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSort}-[func]{quick_sort} ### 快速排序類別 ###
def quick_sort(nums, left, right)
# 子陣列長度不為 1 時遞迴
if left < right
# 哨兵劃分
pivot = partition(nums, left, right)
# 遞迴左子陣列、右子陣列
quick_sort(nums, left, pivot - 1)
quick_sort(nums, pivot + 1, right)
end
nums
end
``` ```
=== "Zig" === "Zig"
@ -1067,9 +1095,38 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSortMedian}-[func]{median_three} ### 選取三個候選元素的中位數 ###
def median_three(nums, left, mid, right)
[class]{QuickSortMedian}-[func]{partition} # 選取三個候選元素的中位數
_l, _m, _r = nums[left], nums[mid], nums[right]
# m 在 l 和 r 之間
return mid if (_l <= _m && _m <= _r) || (_r <= _m && _m <= _l)
# l 在 m 和 r 之間
return left if (_m <= _l && _l <= _r) || (_r <= _l && _l <= _m)
return right
end
### 哨兵劃分（三數取中值）###
def partition(nums, left, right)
### 以 nums[left] 為基準數
med = median_three(nums, left, (left + right) / 2, right)
# 將中位數交換至陣列最左斷
nums[left], nums[med] = nums[med], nums[left]
i, j = left, right
while i < j
while i < j && nums[j] >= nums[left]
j -= 1 # 從右向左找首個小於基準數的元素
end
while i < j && nums[i] <= nums[left]
i += 1 # 從左向右找首個大於基準數的元素
end
# 元素交換
nums[i], nums[j] = nums[j], nums[i]
end
# 將基準數交換至兩子陣列的分界線
nums[i], nums[left] = nums[left], nums[i]
i # 返回基準數的索引
end
``` ```
=== "Zig" === "Zig"
@ -1377,7 +1434,22 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="quick_sort.rb" ```ruby title="quick_sort.rb"
[class]{QuickSortTailCall}-[func]{quick_sort} ### 快速排序（尾遞迴最佳化）###
def quick_sort(nums, left, right)
# 子陣列長度不為 1 時遞迴
while left < right
# 哨兵劃分
pivot = partition(nums, left, right)
# 對兩個子陣列中較短的那個執行快速排序
if pivot - left < right - pivot
quick_sort(nums, left, pivot - 1)
left = pivot + 1 # 剩餘未排序區間為 [pivot + 1, right]
else
quick_sort(nums, pivot + 1, right)
right = pivot - 1 # 剩餘未排序區間為 [left, pivot - 1]
end
end
end
``` ```
=== "Zig" === "Zig"

50
zh-Hant/docs/chapter_sorting/radix_sort.md
Unescape View File

@ -677,11 +677,51 @@ $$
=== "Ruby" === "Ruby"
```ruby title="radix_sort.rb" ```ruby title="radix_sort.rb"
[class]{}-[func]{digit} ### 獲取元素 num 的第 k 位，其中 exp = 10^(k-1) ###
def digit(num, exp)
[class]{}-[func]{counting_sort_digit} # 轉入 exp 而非 k 可以避免在此重複執行昂貴的次方計算
(num / exp) % 10
[class]{}-[func]{radix_sort} end
### 計數排序（根據 nums 第 k 位排序）###
def counting_sort_digit(nums, exp)
# 十進位制的位範圍為 0~9 ，因此需要長度為 10 的桶陣列
counter = Array.new(10, 0)
n = nums.length
# 統計 0~9 各數字的出現次數
for i in 0...n
d = digit(nums[i], exp) # 獲取 nums[i] 第 k 位，記為 d
counter[d] += 1 # 統計數字 d 的出現次數
end
# 求前綴和，將“出現個數”轉換為“陣列索引”
(1...10).each { |i| counter[i] += counter[i - 1] }
# 倒序走訪，根據桶內統計結果，將各元素填入 res
res = Array.new(n, 0)
for i in (n - 1).downto(0)
d = digit(nums[i], exp)
j = counter[d] - 1 # 獲取 d 在陣列中的索引 j
res[j] = nums[i] # 將當前元素填入索引 j
counter[d] -= 1 # 將 d 的數量減 1
end
# 使用結果覆蓋原陣列 nums
(0...n).each { |i| nums[i] = res[i] }
end
### 基數排序 ###
def radix_sort(nums)
# 獲取陣列的最大元素，用於判斷最大位數
m = nums.max
# 按照從低位到高位的順序走訪
exp = 1
while exp <= m
# 對陣列元素的第 k 位執行計數排序
# k = 1 -> exp = 1
# k = 2 -> exp = 10
# 即 exp = 10^(k-1)
counting_sort_digit(nums, exp)
exp *= 10
end
end
``` ```
=== "Zig" === "Zig"

17
zh-Hant/docs/chapter_sorting/selection_sort.md
Unescape View File

@ -306,7 +306,22 @@ comments: true
=== "Ruby" === "Ruby"
```ruby title="selection_sort.rb" ```ruby title="selection_sort.rb"
[class]{}-[func]{selection_sort} ### 選擇排序 ###
def selection_sort(nums)
n = nums.length
# 外迴圈：未排序區間為 [i, n-1]
for i in 0...(n - 1)
# 內迴圈：找到未排序區間內的最小元素
k = i
for j in (i + 1)...n
if nums[j] < nums[k]
k = j # 記錄最小元素的索引
end
end
# 將該最小元素與未排序區間的首個元素交換
nums[i], nums[k] = nums[k], nums[i]
end
end
``` ```
=== "Zig" === "Zig"

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