Bug fixes and improvements (#1380)

* preorder, inorder, postorder -> pre-order, in-order, post-order

* Bug fixes

* Bug fixes

* Update what_is_dsa.md

* Sync zh and zh-hant versions

* Sync zh and zh-hant versions.

* Update performance_evaluation.md and time_complexity.md

* Add @khoaxuantu to the landing page.

* Sync zh and zh-hant versions

* Add @ khoaxuantu to the landing page of zh-hant and en versions.

codes/c/chapter_stack_and_queue/array_stack.c

@@ -90,11 +90,11 @@ int main() {
 
     /* 获取栈的长度 */
     int size = stack->size;
-    printf("栈的长度 size =  %d\n", size);
+    printf("栈的长度 size = %d\n", size);
 
     /* 判断是否为空 */
     bool empty = isEmpty(stack);
-    printf("栈是否为空 = %stack\n", empty ? "true" : "false");
+    printf("栈是否为空 = %s\n", empty ? "true" : "false");
 
     // 释放内存
     delArrayStack(stack);

codes/kotlin/chapter_sorting/merge_sort.kt

@@ -19,7 +19,7 @@ fun merge(nums: IntArray, left: Int, mid: Int, right: Int) {
     while (i <= mid && j <= right) {
         if (nums[i] <= nums[j])
             tmp[k++] = nums[i++]
-        else 
+        else
             tmp[k++] = nums[j++]
     }
     // 将左子数组和右子数组的剩余元素复制到临时数组中

codes/python/chapter_computational_complexity/time_complexity.py

@@ -97,7 +97,9 @@ def linear_log_recur(n: int) -> int:
     """线性对数阶"""
     if n <= 1:
         return 1
-    count: int = linear_log_recur(n // 2) + linear_log_recur(n // 2)
+    # 一分为二，子问题的规模减小一半
+    count = linear_log_recur(n // 2) + linear_log_recur(n // 2)
+    # 当前子问题包含 n 个操作
     for _ in range(n):
         count += 1
     return count
@@ -120,32 +122,32 @@ if __name__ == "__main__":
     n = 8
     print("输入数据大小 n =", n)
 
-    count: int = constant(n)
+    count = constant(n)
     print("常数阶的操作数量 =", count)
 
-    count: int = linear(n)
+    count = linear(n)
     print("线性阶的操作数量 =", count)
-    count: int = array_traversal([0] * n)
+    count = array_traversal([0] * n)
     print("线性阶（遍历数组）的操作数量 =", count)
 
-    count: int = quadratic(n)
+    count = quadratic(n)
     print("平方阶的操作数量 =", count)
     nums = [i for i in range(n, 0, -1)]  # [n, n-1, ..., 2, 1]
-    count: int = bubble_sort(nums)
+    count = bubble_sort(nums)
     print("平方阶（冒泡排序）的操作数量 =", count)
 
-    count: int = exponential(n)
+    count = exponential(n)
     print("指数阶（循环实现）的操作数量 =", count)
-    count: int = exp_recur(n)
+    count = exp_recur(n)
     print("指数阶（递归实现）的操作数量 =", count)
 
-    count: int = logarithmic(n)
+    count = logarithmic(n)
     print("对数阶（循环实现）的操作数量 =", count)
-    count: int = log_recur(n)
+    count = log_recur(n)
     print("对数阶（递归实现）的操作数量 =", count)
 
-    count: int = linear_log_recur(n)
+    count = linear_log_recur(n)
     print("线性对数阶（递归实现）的操作数量 =", count)
 
-    count: int = factorial_recur(n)
+    count = factorial_recur(n)
     print("阶乘阶（递归实现）的操作数量 =", count)

codes/ruby/chapter_dynamic_programming/climbing_stairs_constraint_dp.rb

@@ -5,7 +5,7 @@ Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
 =end
 
 ### 带约束爬楼梯：动态规划 ###
-def climbing_stairs_backtrack(n)
+def climbing_stairs_constraint_dp(n)
   return 1 if n == 1 || n == 2
 
   # 初始化 dp 表，用于存储子问题的解
@@ -26,6 +26,6 @@ end
 if __FILE__ == $0
   n = 9
 
-  res = climbing_stairs_backtrack(n)
+  res = climbing_stairs_constraint_dp(n)
   puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
 end