refactor: review Swift codes for chapter_computational_complexity art… (#396)
* refactor: review Swift codes for chapter_computational_complexity articles * Update time_complexity.swift * Update time_complexity.swift --------- Co-authored-by: Yudong Jin <krahets@163.com>
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@ -33,6 +33,7 @@ func twoSumHashTable(nums: [Int], target: Int) -> [Int] {
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@main
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enum LeetcodeTwoSum {
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/* Driver Code */
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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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@ -88,7 +88,7 @@ func expRecur(n: Int) -> Int {
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}
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/* 对数阶(循环实现) */
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func logarithmic(n: Int) -> Int {
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func logarithmic(n: Double) -> Int {
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var count = 0
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var n = n
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while n > 1 {
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@ -99,7 +99,7 @@ func logarithmic(n: Int) -> Int {
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}
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/* 对数阶(递归实现) */
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func logRecur(n: Int) -> Int {
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func logRecur(n: Double) -> Int {
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if n <= 1 {
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return 0
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}
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@ -112,7 +112,7 @@ func linearLogRecur(n: Double) -> Int {
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return 1
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}
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var count = linearLogRecur(n: n / 2) + linearLogRecur(n: n / 2)
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for _ in 0 ..< Int(n) {
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for _ in sequence(first: 0, next: { $0 < n - 1 ? $0 + 1 : nil }) {
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count += 1
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}
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return count
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@ -158,9 +158,9 @@ enum TimeComplexity {
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count = expRecur(n: n)
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print("指数阶(递归实现)的计算操作数量 = \(count)")
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count = logarithmic(n: n)
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count = logarithmic(n: Double(n))
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print("对数阶(循环实现)的计算操作数量 = \(count)")
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count = logRecur(n: n)
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count = logRecur(n: Double(n))
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print("对数阶(递归实现)的计算操作数量 = \(count)")
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count = linearLogRecur(n: Double(n))
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@ -138,7 +138,7 @@ $$
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```swift title=""
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// 在某运行平台下
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func algorithm(_ n: Int) {
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func algorithm(n: Int) {
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var a = 2 // 1 ns
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a = a + 1 // 1 ns
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a = a * 2 // 10 ns
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@ -340,19 +340,19 @@ $$
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```swift title=""
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// 算法 A 时间复杂度:常数阶
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func algorithmA(_ n: Int) {
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func algorithmA(n: Int) {
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print(0)
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}
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// 算法 B 时间复杂度:线性阶
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func algorithmB(_ n: Int) {
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func algorithmB(n: Int) {
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for _ in 0 ..< n {
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print(0)
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}
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}
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// 算法 C 时间复杂度:常数阶
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func algorithmC(_ n: Int) {
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func algorithmC(n: Int) {
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for _ in 0 ..< 1000000 {
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print(0)
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}
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