Updated files: .sync.cmd src/main/java/cn/whaifree/MonotoneStack/LeetCode739.java src/main/java/cn/whaifree/leetCode/LeetCode/LeetCode42.java

.sync.cmd

@@ -1,5 +1,30 @@
-git add .
-git commit -m "Sync"
-git push origin master
 
+@echo off
+
+setlocal enabledelayedexpansion
+
+git add .
+
+:: 获取暂存区中的更改文件列表
+for /f "delims=" %%a in ('git diff --cached --name-only') do (
+    set "changed_files=!changed_files! %%a"
+)
+
+:: 如果没有文件更改，则设置一个默认的提交信息
+if "%changed_files%"=="" (
+    set "commit_message=No files changed"
+) else (
+    :: 设置提交信息，包括修改的文件列表
+    set "commit_message=Updated files: %changed_files%"
+)
+
+:: 执行git commit
+git commit -m "%commit_message%"
+echo "git push origin master"
+git push origin master
+echo "git push gitee master"
 git push gitee master
+echo "git push github master"
+git push github master
+
+endlocal

src/main/java/cn/whaifree/MonotoneStack/LeetCode739.java

@@ -0,0 +1,59 @@
+package cn.whaifree.MonotoneStack;
+
+import org.junit.Test;
+
+import java.util.Arrays;
+import java.util.Deque;
+import java.util.LinkedList;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/8 11:48
+ * @注释
+ */
+public class LeetCode739 {
+
+    @Test
+    public void test()
+    {
+        int[] temperatures = {73,74,75,71,69,72,76,73};
+        Solution solution = new Solution();
+        int[] res = solution.dailyTemperatures(temperatures);
+        System.out.println(Arrays.toString(res));
+
+    }
+
+    class Solution {
+        /**
+         * 单调栈
+         *
+         * 不断把i放入栈
+         * if nums[i] > stack.peek
+         *      pop = stack.pop
+         *      res[] = i - pop
+         * else
+         *      stack.push(i)
+         *
+         * @param temperatures
+         * @return
+         */
+        public int[] dailyTemperatures(int[] temperatures) {
+            int[] res = new int[temperatures.length];
+            Deque<Integer> stack = new LinkedList<>();
+            stack.push(0);
+            for (int i = 1; i < temperatures.length ; i++) {
+                if (temperatures[i] > temperatures[stack.peek()]) {
+                    // 如果当前比遍历元素比 栈顶的大，证明该元素就是栈顶元素右边第一个大于他的值
+                    // 并且需要判断一下，当前元素是不是 新栈顶 的下一个大于他的
+                    while (!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {
+                        res[stack.peek()] = i - stack.peek();
+                        stack.pop();
+                    }
+                }
+                stack.push(i);
+            }
+            return res;
+        }
+    }
+}

src/main/java/cn/whaifree/leetCode/LeetCode/LeetCode42.java

@@ -0,0 +1,106 @@
+package cn.whaifree.leetCode.LeetCode;
+
+import org.junit.Test;
+
+import java.util.Deque;
+import java.util.LinkedList;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/8 11:23
+ * @注释
+ */
+public class LeetCode42 {
+
+    @Test
+    public void test()
+    {
+
+        int[] height = new int[]{0,1,0,2,1,0,1,3,2,1,2,1};
+        int trap = new Solution2().trap(height);
+        System.out.println(trap);
+    }
+
+    class Solution {
+        /**
+         * 从左到右 获得每个位置的左边最高
+         * 从右到左 获得每个位置的右边最高
+         *
+         * 取交集
+         *
+         * @param height
+         * @return
+         */
+        public int trap(int[] height) {
+            int[] leftHeight = new int[height.length];
+            int leftH = 0;
+            for (int i = 0; i < height.length; i++) {
+                leftH = Math.max(leftH, height[i]);
+                leftHeight[i] = leftH;
+            }
+            int[] rightHeight = new int[height.length];
+            int rightH = 0;
+            for (int i = height.length - 1; i >= 0; i--) {
+                rightH = Math.max(rightH, height[i]);
+                rightHeight[i] = rightH;
+            }
+
+            int res = 0;
+            for (int i = 0; i < height.length; i++) {
+                res += Math.min(rightHeight[i], leftHeight[i]) - height[i];
+            }
+            return res;
+        }
+    }
+
+    class Solution2{
+        /**
+         * 关键在于
+         * - i值 < 栈顶值 入栈
+         *      【3 2 1】 递减，下一个遇到5时，先计算left=2 mid=1 right=5 的容积,再while left=3 mid=1 right=5的容积
+         * - i值 > 栈顶值 需要计算i为槽时的容积
+         * - i值 == 栈顶值 更新
+         * @param height
+         * @return
+         */
+        public int trap(int[] height) {
+            // 单调栈 找到右边第一个高于他的墙
+            // 保证栈内元素，从上到下为递增的，这样就能获取到槽的两边，并且确保不会有一边为空的
+            Deque<Integer> stack = new LinkedList<>();
+            stack.push(0);
+            int res = 0;
+            for (int i = 1; i < height.length; i++) {
+                if (height[i] > height[stack.peek()]) {
+                    // 遇到凹陷
+                    //pop up all lower value
+                    int heightAtIdx = height[i]; // 右边墙的高度
+                    // 不断判断左边的墙，因为如果左边的墙存在，必然大于上一个左墙   如height 3 2 1 5
+                    while (!stack.isEmpty() && (heightAtIdx > height[stack.peek()])){
+                        int mid = stack.pop(); // 中间槽的下标
+
+                        if (!stack.isEmpty()){
+                            int left = stack.peek(); // 左边墙的下标
+
+                            int h = Math.min(height[left], height[i]) - height[mid];
+                            int w = i - left - 1;
+                            int hold = h * w;
+
+                            if (hold > 0) res += hold;
+                        }
+                    }
+                    stack.push(i);
+
+
+                } else if (height[i] == height[stack.peek()]) {
+                    stack.pop();
+                    stack.push(i);
+                }else {
+                    stack.push(i);
+                }
+            }
+
+            return res;
+        }
+    }
+}