接雨水

问题描述：

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.
如上图所示，海拔分别为 [0,1,0,2,1,0,1,3,2,1,2,1], 返回 6.

挑战

O(n) 时间, O(1) 空间

思路：题目不难，观察下就可以发现被水填满后的形状是先升后降的塔形，因此，先遍历一遍找到塔顶，然后分别从两边开始，往塔顶所在位置遍历，水位只会增高不会减小，且一直和最近遇到的最大高度持平，这样知道了实时水位，就可以边遍历边计算面积。
c++代码如下：

class Solution {public:    /*     * @param heights: a list of integers     * @return: a integer     */    int trapRainWater(vector<int> &heights) {        // write your code here        //防止越界        if(heights.size() < 2)return 0;        int maxheight = 0;        int maxindex = 0;        //寻找最高峰        for(int i = 0;i!=heights.size();++i){            if(heights[i]>maxheight){                maxheight = heights[i];                maxindex = i;            }        }        int waterarea = 0;        int pedge = 0;        int qedge = 0;        for(;qedge<maxindex;){            qedge++;            if(heights[qedge] > heights[pedge]){                pedge = qedge;            }            else{                waterarea = waterarea+heights[pedge]-heights[qedge];            }        }        pedge = heights.size() - 1;        qedge = heights.size() - 1;        for(;qedge>maxindex;){            qedge--;            if(heights[qedge] > heights[pedge]){                pedge = qedge;            }            else{                waterarea = waterarea+heights[pedge]-heights[qedge];            }        }        return waterarea;    }};