build

docs-en/chapter_computational_complexity/iteration_and_recursion.md

@@ -1631,7 +1631,7 @@ Therefore, **we can use an explicit stack to simulate the behavior of the call s
         var stack: [Int] = []
         var res = 0
         // 递：递归调用
-        for i in stride(from: n, to: 0, by: -1) {
+        for i in (1 ... n).reversed() {
             // 通过“入栈操作”模拟“递”
             stack.append(i)
         }

docs-en/chapter_computational_complexity/time_complexity.md

@@ -321,7 +321,7 @@ Let's understand this concept of "time growth trend" with an example. Assume the
 
     // Time complexity of algorithm C: constant order
     func algorithmC(n: Int) {
-        for _ in 0 ..< 1000000 {
+        for _ in 0 ..< 1_000_000 {
             print(0)
         }
     }
@@ -1780,7 +1780,7 @@ For instance, in bubble sort, the outer loop runs $n - 1$ times, and the inner l
     func bubbleSort(nums: inout [Int]) -> Int {
         var count = 0 // 计数器
         // 外循环：未排序区间为 [0, i]
-        for i in stride(from: nums.count - 1, to: 0, by: -1) {
+        for i in nums.indices.dropFirst().reversed() {
             // 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] {