LeetCode--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!

思路：双指针法。找到最高的柱子，然后从两边分别扫描，从左往右更新最高的柱子，用最高的柱子减去当前的柱子高度就是盛的水，从右往左同理。

class Solution {public:    int trap(vector<int>& height) {        int result=0;        int max=0;        for(int i=0;i<height.size();i++)        {            if(height[i]>height[max])                max=i;        }        for(int left=0,peak=0;left<max;left++)        {            if(height[left]>peak)                peak=height[left];            result+=peak-height[left];        }        for(int right=height.size()-1,top=0;right>max;right--)        {            if(height[right]>top)                top=height[right];            result+=top-height[right];        }        return result;    }};