Merge branch 'krahets:main' into yuelinxin-en

Author: yuelinxin

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/go/chapter_stack_and_queue/array_deque.go

@@ -72,6 +72,9 @@ func (q *arrayDeque) pushLast(num int) {
 /* 队首出队 */
 func (q *arrayDeque) popFirst() any {
 	num := q.peekFirst()
+	if num == nil {
+		return nil
+	}
 	// 队首指针向后移动一位
 	q.front = q.index(q.front + 1)
 	q.queSize--
@@ -81,6 +84,9 @@ func (q *arrayDeque) popFirst() any {
 /* 队尾出队 */
 func (q *arrayDeque) popLast() any {
 	num := q.peekLast()
+	if num == nil {
+		return nil
+	}
 	q.queSize--
 	return num
 }

codes/go/chapter_stack_and_queue/array_queue.go

@@ -49,6 +49,10 @@ func (q *arrayQueue) push(num int) {
 /* 出队 */
 func (q *arrayQueue) pop() any {
 	num := q.peek()
+	if num == nil {
+		return nil
+	}
+
 	// 队首指针向后移动一位，若越过尾部，则返回到数组头部
 	q.front = (q.front + 1) % q.queCapacity
 	q.queSize--

codes/go/chapter_stack_and_queue/queue_test.go

@@ -46,9 +46,13 @@ func TestQueue(t *testing.T) {
 }
 
 func TestArrayQueue(t *testing.T) {
+
 	// 初始化队列，使用队列的通用接口
 	capacity := 10
 	queue := newArrayQueue(capacity)
+	if queue.pop() != nil {
+		t.Errorf("want:%v,got:%v", nil, queue.pop())
+	}
 
 	// 元素入队
 	queue.push(1)

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 @@ def factorial_recur(n: int) -> int:
     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_backtrack.rb

@@ -0,0 +1,37 @@
+=begin
+File: climbing_stairs_backtrack.rb
+Created Time: 2024-05-29
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### 回溯 ###
+def backtrack(choices, state, n, res)
+  # 当爬到第 n 阶时，方案数量加 1
+  res[0] += 1 if state == n
+  # 遍历所有选择
+  for choice in choices
+    # 剪枝：不允许越过第 n 阶
+    next if state + choice > n
+
+    # 尝试：做出选择，更新状态
+    backtrack(choices, state + choice, n, res)
+  end
+  # 回退
+end
+
+### 爬楼梯：回溯 ###
+def climbing_stairs_backtrack(n)
+  choices = [1, 2] # 可选择向上爬 1 阶或 2 阶
+  state = 0 # 从第 0 阶开始爬
+  res = [0] # 使用 res[0] 记录方案数量
+  backtrack(choices, state, n, res)
+  res.first
+end
+
+### Driver Code ###
+if __FILE__ == $0
+  n = 9
+
+  res = climbing_stairs_backtrack(n)
+  puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
+end

codes/ruby/chapter_dynamic_programming/climbing_stairs_constraint_dp.rb

@@ -0,0 +1,31 @@
+=begin
+File: climbing_stairs_constraint_dp.rb
+Created Time: 2024-05-29
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### 带约束爬楼梯：动态规划 ###
+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
+  dp = Array.new(n + 1) { Array.new(3, 0) }
+  # 初始状态：预设最小子问题的解
+  dp[1][1], dp[1][2] = 1, 0
+  dp[2][1], dp[2][2] = 0, 1
+  # 状态转移：从较小子问题逐步求解较大子问题
+  for i in 3...(n + 1)
+    dp[i][1] = dp[i - 1][2]
+    dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
+  end
+
+  dp[n][1] + dp[n][2]
+end
+
+### Driver Code ###
+if __FILE__ == $0
+  n = 9
+
+  res = climbing_stairs_constraint_dp(n)
+  puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
+end

codes/ruby/chapter_dynamic_programming/climbing_stairs_dfs.rb

@@ -0,0 +1,26 @@
+=begin
+File: climbing_stairs_dfs.rb
+Created Time: 2024-05-29
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### 搜索 ###
+def dfs(i)
+  # 已知 dp[1] 和 dp[2] ，返回之
+  return i if i == 1 || i == 2
+  # dp[i] = dp[i-1] + dp[i-2]
+  dfs(i - 1) + dfs(i - 2)
+end
+
+### 爬楼梯：搜索 ###
+def climbing_stairs_dfs(n)
+  dfs(n)
+end
+
+### Driver Code ###
+if __FILE__ == $0
+  n = 9
+
+  res = climbing_stairs_dfs(n)
+  puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
+end

codes/ruby/chapter_dynamic_programming/climbing_stairs_dfs_mem.rb

@@ -0,0 +1,33 @@
+=begin
+File: climbing_stairs_dfs_mem.rb
+Created Time: 2024-05-29
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### 记忆化搜索 ###
+def dfs(i, mem)
+  # 已知 dp[1] 和 dp[2] ，返回之
+  return i if i == 1 || i == 2
+  # 若存在记录 dp[i] ，则直接返回之
+  return mem[i] if mem[i] != -1
+
+  # dp[i] = dp[i-1] + dp[i-2]
+  count = dfs(i - 1, mem) + dfs(i - 2, mem)
+  # 记录 dp[i]
+  mem[i] = count
+end
+
+### 爬楼梯：记忆化搜索 ###
+def climbing_stairs_dfs_mem(n)
+  # mem[i] 记录爬到第 i 阶的方案总数，-1 代表无记录
+  mem = Array.new(n + 1, -1)
+  dfs(n, mem)
+end
+
+### Driver Code ###
+if __FILE__ == $0
+  n = 9
+
+  res = climbing_stairs_dfs_mem(n)
+  puts "爬 #{n} 阶楼梯共有 #{res} 种方案"
+end