LeetCode 042 Trapping Rain Water

浏览数：24 / 时间：2015年06月11日

题目要求：Trapping Rain Water

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.

技术分享

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!

代码如下：

class Solution {
public:
    int trap(int A[], int n) {
        
        int maxIdx = 0;
        int water = 0;
        
        //找到最长的木板，设为maxIdx
        for(int i = 1; i < n; i++){
            if(A[i] > A[maxIdx]){
                maxIdx = i;
            }
        }
        
        int max = A[0];
        
        //左侧逼近
        for(int i = 1; i < maxIdx; i++){
            
            if(max < A[i]) 
                max = A[i];
            //木板左边(max)和右边(最高)都比它高，则可以放
            // max - 该木板长度 的水
            else 
                water += max - A[i];
        }
        
        //右侧逼近
        max = A[n - 1];
        for(int i = n - 2; i >= maxIdx; i--){
            if(max < A[i]) max = A[i];
            else water += max - A[i];
        }
        
        return water;
        
    }
};