LeetCode - Trapping Rain Water

最新推荐文章于 2025-09-10 15:24:29 发布

weixin_30823001

最新推荐文章于 2025-09-10 15:24:29 发布

阅读量45

点赞数

CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/clover-xuqi/p/7154799.html

本文介绍了一种解决雨水收集问题的方法。通过计算给定高度数组中可以存储的最大水量，使用两次遍历算法实现。示例中，对于数组[0,1,0,2,1,0,1,3,2,1,2,1]，能存储6单位的雨水。

题目描述：

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!

AC代码：

int right(int* a,int n){
    int ans = 0;
    int has = 0;
    int h = 0;
    for(int i = 0 ; i < n ; i ++){
        if(a[i]>=h){
            ans+=has;
            h=a[i];
            has = 0;
        }else{
            has+=h-a[i];
        }
    }
    return ans;
}

int left(int*a,int n){
    int ans = 0;
    int has = 0;
    int h = 0;
    for(int i = n-1 ; i >= 0; i --){
        if(a[i]>h){
            ans+=has;
            h = a[i];
            has = 0;
        }else{
            has += h-a[i];
        }
    }
    return ans;
}