feat: add Swift codes for space time tradeoff article

pull/205/head
nuomi1 2 years ago
parent 4ac254d1f7
commit 2069dddd3b
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GPG Key ID: E410D5FF602FBF25

@ -8,11 +8,13 @@ let package = Package(
.executable(name: "time_complexity", targets: ["time_complexity"]),
.executable(name: "worst_best_time_complexity", targets: ["worst_best_time_complexity"]),
.executable(name: "space_complexity", targets: ["space_complexity"]),
.executable(name: "leetcode_two_sum", targets: ["leetcode_two_sum"]),
],
targets: [
.target(name: "utils", path: "utils"),
.executableTarget(name: "time_complexity", path: "chapter_computational_complexity", sources: ["time_complexity.swift"]),
.executableTarget(name: "worst_best_time_complexity", path: "chapter_computational_complexity", sources: ["worst_best_time_complexity.swift"]),
.executableTarget(name: "space_complexity", dependencies: ["utils"], path: "chapter_computational_complexity", sources: ["space_complexity.swift"]),
.executableTarget(name: "leetcode_two_sum", path: "chapter_computational_complexity", sources: ["leetcode_two_sum.swift"]),
]
)

@ -0,0 +1,46 @@
/*
* File: leetcode_two_sum.swift
* Created Time: 2023-01-03
* Author: nuomi1 (nuomi1@qq.com)
*/
func twoSumBruteForce(nums: [Int], target: Int) -> [Int] {
// O(n^2)
for i in nums.indices.dropLast() {
for j in nums.indices.dropFirst(i + 1) {
if nums[i] + nums[j] == target {
return [i, j]
}
}
}
return [0]
}
func twoSumHashTable(nums: [Int], target: Int) -> [Int] {
// O(n)
var dic: [Int: Int] = [:]
// O(n)
for i in nums.indices {
if let j = dic[target - nums[i]] {
return [j, i]
}
dic[nums[i]] = i
}
return [0]
}
@main
enum LeetcodeTwoSum {
static func main() {
// ======= Test Case =======
let nums = [2, 7, 11, 15]
let target = 9
// ====== Driver Code ======
//
var res = twoSumBruteForce(nums: nums, target: target)
print("方法一 res = \(res)")
//
res = twoSumHashTable(nums: nums, target: target)
print("方法二 res = \(res)")
}
}

@ -149,6 +149,22 @@ comments: true
}
```
=== "Swift"
```swift title="leetcode_two_sum.swift"
func twoSumBruteForce(nums: [Int], target: Int) -> [Int] {
// 两层循环,时间复杂度 O(n^2)
for i in nums.indices.dropLast() {
for j in nums.indices.dropFirst(i + 1) {
if nums[i] + nums[j] == target {
return [i, j]
}
}
}
return [0]
}
```
### 方法二:辅助哈希表
时间复杂度 $O(N)$ ,空间复杂度 $O(N)$ ,属于「空间换时间」。
@ -294,3 +310,20 @@ comments: true
}
}
```
=== "Swift"
```swift title="leetcode_two_sum.swift"
func twoSumHashTable(nums: [Int], target: Int) -> [Int] {
// 辅助哈希表,空间复杂度 O(n)
var dic: [Int: Int] = [:]
// 单层循环,时间复杂度 O(n)
for i in nums.indices {
if let j = dic[target - nums[i]] {
return [j, i]
}
dic[nums[i]] = i
}
return [0]
}
```

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