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feat: add Swift codes for space time tradeoff article
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@ -8,11 +8,13 @@ let package = Package(
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.executable(name: "time_complexity", targets: ["time_complexity"]),
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.executable(name: "worst_best_time_complexity", targets: ["worst_best_time_complexity"]),
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.executable(name: "space_complexity", targets: ["space_complexity"]),
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.executable(name: "leetcode_two_sum", targets: ["leetcode_two_sum"]),
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],
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targets: [
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.target(name: "utils", path: "utils"),
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.executableTarget(name: "time_complexity", path: "chapter_computational_complexity", sources: ["time_complexity.swift"]),
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.executableTarget(name: "worst_best_time_complexity", path: "chapter_computational_complexity", sources: ["worst_best_time_complexity.swift"]),
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.executableTarget(name: "space_complexity", dependencies: ["utils"], path: "chapter_computational_complexity", sources: ["space_complexity.swift"]),
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.executableTarget(name: "leetcode_two_sum", path: "chapter_computational_complexity", sources: ["leetcode_two_sum.swift"]),
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]
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)
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@ -0,0 +1,46 @@
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/*
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* File: leetcode_two_sum.swift
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* Created Time: 2023-01-03
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* Author: nuomi1 (nuomi1@qq.com)
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*/
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func twoSumBruteForce(nums: [Int], target: Int) -> [Int] {
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// 两层循环,时间复杂度 O(n^2)
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for i in nums.indices.dropLast() {
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for j in nums.indices.dropFirst(i + 1) {
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if nums[i] + nums[j] == target {
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return [i, j]
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}
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}
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}
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return [0]
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}
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func twoSumHashTable(nums: [Int], target: Int) -> [Int] {
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// 辅助哈希表,空间复杂度 O(n)
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var dic: [Int: Int] = [:]
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// 单层循环,时间复杂度 O(n)
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for i in nums.indices {
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if let j = dic[target - nums[i]] {
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return [j, i]
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}
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dic[nums[i]] = i
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}
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return [0]
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}
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@main
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enum LeetcodeTwoSum {
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static func main() {
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// ======= Test Case =======
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let nums = [2, 7, 11, 15]
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let target = 9
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// ====== Driver Code ======
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// 方法一
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var res = twoSumBruteForce(nums: nums, target: target)
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print("方法一 res = \(res)")
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// 方法二
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res = twoSumHashTable(nums: nums, target: target)
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print("方法二 res = \(res)")
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}
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}
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@ -149,6 +149,22 @@ comments: true
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}
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```
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=== "Swift"
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```swift title="leetcode_two_sum.swift"
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func twoSumBruteForce(nums: [Int], target: Int) -> [Int] {
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// 两层循环,时间复杂度 O(n^2)
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for i in nums.indices.dropLast() {
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for j in nums.indices.dropFirst(i + 1) {
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if nums[i] + nums[j] == target {
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return [i, j]
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}
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}
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}
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return [0]
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}
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```
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### 方法二:辅助哈希表
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时间复杂度 $O(N)$ ,空间复杂度 $O(N)$ ,属于「空间换时间」。
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@ -294,3 +310,20 @@ comments: true
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}
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}
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```
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=== "Swift"
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```swift title="leetcode_two_sum.swift"
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func twoSumHashTable(nums: [Int], target: Int) -> [Int] {
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// 辅助哈希表,空间复杂度 O(n)
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var dic: [Int: Int] = [:]
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// 单层循环,时间复杂度 O(n)
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for i in nums.indices {
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if let j = dic[target - nums[i]] {
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return [j, i]
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}
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dic[nums[i]] = i
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}
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return [0]
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}
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```
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