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a/docs/chapter_dynamic_programming/edit_distance_problem.assets/edit_distance_decision_tree.png and b/docs/chapter_dynamic_programming/edit_distance_problem.assets/edit_distance_decision_tree.png differ diff --git a/docs/chapter_dynamic_programming/intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png b/docs/chapter_dynamic_programming/intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png index bd2166a24..6b28c0930 100644 Binary files a/docs/chapter_dynamic_programming/intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png and b/docs/chapter_dynamic_programming/intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png differ diff --git a/docs/chapter_dynamic_programming/summary.md b/docs/chapter_dynamic_programming/summary.md index d6fd3c9a4..5d422a9d4 100644 --- a/docs/chapter_dynamic_programming/summary.md +++ b/docs/chapter_dynamic_programming/summary.md @@ -14,7 +14,7 @@ - 0-1 背包的状态定义为前 $i$ 个物品在剩余容量为 $c$ 的背包中的最大价值。根据不放入背包和放入背包两种决策，可得到最优子结构，并构建出状态转移方程。在空间优化中，由于每个状态依赖正上方和左上方的状态，因此需要倒序遍历列表，避免左上方状态被覆盖。 - 完全背包问题的每种物品的选取数量无限制，因此选择放入物品的状态转移与 0-1 背包问题不同。由于状态依赖正上方和正左方的状态，因此在空间优化中应当正序遍历。 - 零钱兑换问题是完全背包问题的一个变种。它从求“最大”价值变为求“最小”硬币数量，因此状态转移方程中的 $\max()$ 应改为 $\min()$ 。从追求“不超过”背包容量到追求“恰好”凑出目标金额，因此使用 $amt + 1$ 来表示“无法凑出目标金额”的无效解。 -- 零钱兑换 II 问题从求“最少硬币数量”改为求“硬币组合数量”，状态转移方程相应地从 $\min()$ 改为求和运算符。 +- 零钱兑换问题 II 从求“最少硬币数量”改为求“硬币组合数量”，状态转移方程相应地从 $\min()$ 改为求和运算符。 **编辑距离问题** diff --git a/docs/chapter_graph/graph_operations.assets/adjacency_matrix_add_vertex.png b/docs/chapter_graph/graph_operations.assets/adjacency_matrix_add_vertex.png index a76fa8a95..87d03db82 100644 Binary files a/docs/chapter_graph/graph_operations.assets/adjacency_matrix_add_vertex.png and b/docs/chapter_graph/graph_operations.assets/adjacency_matrix_add_vertex.png differ diff --git a/docs/chapter_graph/graph_traversal.assets/graph_bfs.png b/docs/chapter_graph/graph_traversal.assets/graph_bfs.png index cc46b9031..96a68291f 100644 Binary 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b/docs/chapter_greedy/greedy_algorithm.assets/coin_change_greedy_strategy.png index b2daea2c7..1f8b4e8d2 100644 Binary files a/docs/chapter_greedy/greedy_algorithm.assets/coin_change_greedy_strategy.png and b/docs/chapter_greedy/greedy_algorithm.assets/coin_change_greedy_strategy.png differ diff --git a/docs/chapter_greedy/greedy_algorithm.assets/coin_change_greedy_vs_dp.png b/docs/chapter_greedy/greedy_algorithm.assets/coin_change_greedy_vs_dp.png index d7cb513c7..d13ed01a1 100644 Binary files a/docs/chapter_greedy/greedy_algorithm.assets/coin_change_greedy_vs_dp.png and b/docs/chapter_greedy/greedy_algorithm.assets/coin_change_greedy_vs_dp.png differ diff --git a/docs/chapter_greedy/greedy_algorithm.md b/docs/chapter_greedy/greedy_algorithm.md index 2cd435499..05d433d9b 100644 --- a/docs/chapter_greedy/greedy_algorithm.md +++ b/docs/chapter_greedy/greedy_algorithm.md @@ -27,7 +27,7 @@ ## 贪心算法的优点与局限性 -**贪心算法不仅操作直接、实现简单，而且通常效率也很高**。在以上代码中，记硬币最小面值为 $\min(coins)$ ，则贪心选择最多循环 $amt / \min(coins)$ 次，时间复杂度为 $O(amt / \min(coins))$ 。这比动态规划解法的时间复杂度 $O(n \times amt)$ 降低了一个数量级。 +**贪心算法不仅操作直接、实现简单，而且通常效率也很高**。在以上代码中，记硬币最小面值为 $\min(coins)$ ，则贪心选择最多循环 $amt / \min(coins)$ 次，时间复杂度为 $O(amt / \min(coins))$ 。这比动态规划解法的时间复杂度 $O(n \times amt)$ 小了一个数量级。 然而，**对于某些硬币面值组合，贪心算法并不能找到最优解**。下图给出了两个示例。 diff --git a/docs/chapter_greedy/max_capacity_problem.assets/max_capacity_greedy_step1.png b/docs/chapter_greedy/max_capacity_problem.assets/max_capacity_greedy_step1.png index 3e73587d9..9d7f8aa6b 100644 Binary files a/docs/chapter_greedy/max_capacity_problem.assets/max_capacity_greedy_step1.png and b/docs/chapter_greedy/max_capacity_problem.assets/max_capacity_greedy_step1.png differ diff --git a/docs/chapter_hashing/hash_collision.assets/hash_table_chaining.png b/docs/chapter_hashing/hash_collision.assets/hash_table_chaining.png index bc91ce226..1c67feeff 100644 Binary files a/docs/chapter_hashing/hash_collision.assets/hash_table_chaining.png and 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表   双向队列操作效率

-| 方法名 | 描述 | 时间复杂度 | -| ------------- | ---------------- | ---------- | -| `pushFirst()` | 将元素添加至队首 | $O(1)$ | -| `pushLast()` | 将元素添加至队尾 | $O(1)$ | -| `popFirst()` | 删除队首元素 | $O(1)$ | -| `popLast()` | 删除队尾元素 | $O(1)$ | -| `peekFirst()` | 访问队首元素 | $O(1)$ | -| `peekLast()` | 访问队尾元素 | $O(1)$ | +| 方法名 | 描述 | 时间复杂度 | +| -------------- | ---------------- | ---------- | +| `push_first()` | 将元素添加至队首 | $O(1)$ | +| `push_last()` | 将元素添加至队尾 | $O(1)$ | +| `pop_first()` | 删除队首元素 | $O(1)$ | +| `pop_last()` | 删除队尾元素 | $O(1)$ | +| `peek_first()` | 访问队首元素 | $O(1)$ | +| `peek_last()` | 访问队尾元素 | $O(1)$ | 同样地，我们可以直接使用编程语言中已实现的双向队列类： @@ -349,16 +349,16 @@ === "LinkedListDeque" ![基于链表实现双向队列的入队出队操作](deque.assets/linkedlist_deque.png) -=== "pushLast()" +=== "push_last()" ![linkedlist_deque_push_last](deque.assets/linkedlist_deque_push_last.png) -=== "pushFirst()" +=== "push_first()" ![linkedlist_deque_push_first](deque.assets/linkedlist_deque_push_first.png) -=== "popLast()" +=== "pop_last()" ![linkedlist_deque_pop_last](deque.assets/linkedlist_deque_pop_last.png) -=== "popFirst()" +=== "pop_first()" ![linkedlist_deque_pop_first](deque.assets/linkedlist_deque_pop_first.png) 实现代码如下所示： @@ -374,16 +374,16 @@ === "ArrayDeque" ![基于数组实现双向队列的入队出队操作](deque.assets/array_deque.png) -=== "pushLast()" +=== "push_last()" ![array_deque_push_last](deque.assets/array_deque_push_last.png) -=== "pushFirst()" +=== "push_first()" ![array_deque_push_first](deque.assets/array_deque_push_first.png) -=== "popLast()" +=== "pop_last()" ![array_deque_pop_last](deque.assets/array_deque_pop_last.png) -=== "popFirst()" +=== "pop_first()" ![array_deque_pop_first](deque.assets/array_deque_pop_first.png) 在队列的实现基础上，仅需增加“队首入队”和“队尾出队”的方法： diff --git a/docs/chapter_stack_and_queue/queue.assets/array_queue.png b/docs/chapter_stack_and_queue/queue.assets/array_queue.png index 29c7f4962..2c6b92a65 100644 Binary files a/docs/chapter_stack_and_queue/queue.assets/array_queue.png and 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- - + +

《 Hello 算法 》

@@ -134,7 +134,7 @@ hide: 两年前，我在力扣上分享了“剑指 Offer”系列题解，受到了许多同学的喜爱和支持。在与读者交流期间，我最常收到的一个问题是“如何入门算法”。逐渐地，我对这个问题产生了浓厚的兴趣。 -两眼一抹黑地刷题似乎是最受欢迎的方法，简单直接且有效。然而刷题就如同玩“扫雷”游戏，自学能力强的同学能够顺利将地雷逐个排掉，而基础不足的同学很可能被炸的满头是包，并在挫折中步步退缩。通读教材也是一种常见做法，但对于面向求职的同学来说，毕业季、投递简历、准备笔试面试已经消耗了大部分精力，啃厚重的书往往变成了一项艰巨的挑战。 +两眼一抹黑地刷题似乎是最受欢迎的方法，简单直接且有效。然而刷题就如同玩“扫雷”游戏，自学能力强的同学能够顺利将地雷逐个排掉，而基础不足的同学很可能被炸得满头是包，并在挫折中步步退缩。通读教材也是一种常见做法，但对于面向求职的同学来说，毕业季、投递简历、准备笔试面试已经消耗了大部分精力，啃厚重的书往往变成了一项艰巨的挑战。 如果你也面临类似的困扰，那么很幸运这本书找到了你。本书是我对这个问题给出的答案，即使不是最优解，也至少是一次积极的尝试。本书虽然不足以让你直接拿到 Offer ，但会引导你探索数据结构与算法的“知识地图”，带你了解不同“地雷”的形状、大小和分布位置，让你掌握各种“排雷方法”。有了这些本领，相信你可以更加自如地刷题和阅读文献，逐步构建起完整的知识体系。 diff --git a/mkdocs-en.yml b/mkdocs-en.yml index c86f9cff8..6fd14b9a5 100644 --- a/mkdocs-en.yml +++ b/mkdocs-en.yml @@ -16,7 +16,7 @@ theme: language: en font: text: Roboto - code: Fira Code + code: Roboto Mono extra: status: