leetcode -- Trapping Rain Water

最新推荐文章于 2025-12-17 19:02:04 发布

转载最新推荐文章于 2025-12-17 19:02:04 发布 · 57 阅读

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#java

本文详细解析了如何通过计算特定高度图中能蓄积多少雨水来解决经典算法问题。使用Java实现，通过寻找每个位置的最大左右边界来确定该位置可以蓄积的雨水量。

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.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

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!

本题寻找当前index的最高左右边界，当前index可蓄水量=(min(left, right) - cur)

 1 public class Solution {
 2     public int trap(int[] A) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int len = A.length;
 6         int l = 0, r = 0;
 7         int result = 0;
 8         
 9         for(int i = 1; i < len - 1; i++){
10             for(int m = 0; m < i; m ++){
11                 if(A[m] > l){
12                     l = A[m];
13                 }            
14             }
15             
16             for(int n = len - 1; n > i; n --){
17                 if(A[n] > r){
18                     r = A[n];
19                 }
20             }
21             
22             int min = Math.min(l, r);
23             if(min - A[i] > 0){
24                 result += min - A[i];
25             }
26             l = 0;
27             r = 0;
28         }
29         return result;
30     }
31 }