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[leetcode] Trapping Rain Water

时间:2014-12-20 16:49:47      阅读:159      评论:0      收藏:0      [点我收藏+]

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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

思路:

不会做,详细分析和思路参考此文。看懂了别人的,照葫芦画瓢弄了一个。

题解:

技术分享
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

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原文地址:http://www.cnblogs.com/jiasaidongqi/p/4175567.html

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