LeetCode OJ:Trapping Rain Water

浏览数：19 / 时间：2015年06月11日

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.

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!

算法思想：

有点类似于最大子矩阵问题：

LeetCode OJ:Maximal Rectangle

class Solution {
public:
    int trap(int A[], int n) {
        if(!n)return 0;
        vector<int> L(n);
        vector<int> R(n,n);
        int result=0;
        int maxH=0;
        for(int i=0;i<n;i++){
            L[i]=maxH;
            maxH=max(L[i],A[i]);
        }
        R[n-1]=0;
        maxH=A[n-1];
        for(int i=n-2;i>0;i--){
            R[i]=maxH;
            maxH=max(R[i],A[i]);
            int height=min(L[i],R[i]);
            if(height>A[i]){
                result+=height-A[i];
            }
        }
        return result;
    }
};

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LeetCode OJ:Trapping Rain Water