四刷42

Author: diguage

docs/0042-trapping-rain-water.adoc

@@ -1,24 +1,39 @@
 [#0042-trapping-rain-water]
-= 42. Trapping Rain Water
+= 42. 接雨水
 
-{leetcode}/problems/trapping-rain-water/[LeetCode - Trapping Rain Water^]
+https://leetcode.cn/problems/trapping-rain-water/[LeetCode - 42. 接雨水 ^]
 
-Given _n_ non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
+给定 `n` 个非负整数表示每个宽度为 `1` 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。
 
-*Example:*
+*示例 1：*
 
 image::images/0042-00.png[{image_attr}]
 
-[.small]#The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. *Thanks Marcos* for contributing this image!#
+....
+输入：height = [0,1,0,2,1,0,1,3,2,1,2,1]
+输出：6
+解释：上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。
+....
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* [0,1,0,2,1,0,1,3,2,1,2,1]
-*Output:* 6
-----
+*示例 2：*
+
+....
+输入：height = [4,2,0,3,2,5]
+输出：9
+....
+
+*提示：*
 
+* `n == height.length`
+* `1 \<= n \<= 2 * 10^4^`
+* `0 \<= height[i] \<= 10^5^`
+
+
+[#解题分析]
 == 解题分析
 
+单调栈解法图示：
+
 image::images/0042-01.png[{image_attr}]
 
 image::images/0042-02.png[{image_attr}]
@@ -61,6 +76,15 @@ include::{sourcedir}/_0042_TrappingRainWater_2.java[tag=answer]
 include::{sourcedir}/_0042_TrappingRainWater_3.java[tag=answer]
 ----
 --
+
+四刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0042_TrappingRainWater_4.java[tag=answer]
+----
+--
 ====
 
 
@@ -70,6 +94,7 @@ include::{sourcedir}/_0042_TrappingRainWater_3.java[tag=answer]
 
 == 参考资料
 
+. https://leetcode.cn/problems/trapping-rain-water/solutions/692342/jie-yu-shui-by-leetcode-solution-tuvc/[42. 接雨水 - 官方题解^] -- 左右双指针的解法，代码最简单。
 . https://leetcode.cn/problems/trapping-rain-water/solutions/9112/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-w-8/[42. 接雨水 - 详细通俗的思路分析，多解法^]
 . https://blog.csdn.net/weixin_50348837/article/details/136304458[深入理解单调栈算法，这一篇就够了^]
 . https://leetcode.cn/problems/trapping-rain-water/solutions/185678/trapping-rain-water-by-ikaruga/[42. 接雨水 - 单调递减栈，简洁代码，动图模拟^]

logbook/202503.adoc

@@ -368,6 +368,11 @@ endif::[]
 |{doc_base_url}/0027-remove-element.adoc[题解]
 |✅ 快慢指针。慢指针记录满足条件的长度，快指针指向需要处理的元素。
 
+|{counter:codes}
+|{leetcode_base_url}/trapping-rain-water/[42. 接雨水^]
+|{doc_base_url}/0042-trapping-rain-water.adoc[题解]
+|❌ 使用单调栈没写出来。使用单调栈，需 `h × w`，不能只看头顶的容量。使用左右双指针夹逼代码更简单，只需要计算头顶的容量即可。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

src/main/java/com/diguage/algo/leetcode/_0042_TrappingRainWater_4.java

@@ -0,0 +1,36 @@
+package com.diguage.algo.leetcode;
+
+import java.util.ArrayDeque;
+import java.util.Deque;
+
+public class _0042_TrappingRainWater_4 {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-04-20 18:38:35
+   */
+  public int trap(int[] height) {
+    Deque<Integer> stack = new ArrayDeque<>(height.length);
+    int result = 0;
+    for (int right = 0; right < height.length; right++) {
+      while (!stack.isEmpty() && height[stack.peek()] < height[right]) {
+        int mid = stack.poll();
+        if (stack.isEmpty()) {
+          break;
+        }
+        Integer left = stack.peek();
+        int h = Math.min(height[left], height[right]) - height[mid];
+        int w = right - left - 1;
+        result += h * w;
+      }
+      stack.push(right);
+    }
+    return result;
+  }
+
+  // end::answer[]
+  public static void main(String[] args) {
+    new _0042_TrappingRainWater_4().trap(new int[]{4, 2, 0, 3, 2, 5});
+  }
+}