[leetcode] Trapping Rain Water

標籤：

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.

思路：

不會做，詳細分析和思路參考此文。看懂了別人的，照葫蘆畫瓢弄了一個。

題解：

class Solution {public:    int trap(int A[], int n) {        int res = 0;        int maxIndex = 0;        for(int i=1;i<n;i++)            if(A[i]>A[maxIndex])                maxIndex = i;        int maxCur = A[0];        for(int i=1;i<maxIndex;i++) {            if(A[i]<maxCur)                res += maxCur-A[i];            else                maxCur = A[i];        }        maxCur = A[n-1];        for(int i=n-2;i>maxIndex;i--) {            if(A[i]<maxCur)                res += maxCur-A[i];            else                maxCur = A[i];        }        return res;    }};

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[leetcode] Trapping Rain Water