Trapping Rain Water

题目：
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：
1、左右两边分别进行扫描，分别找出左右两边的最大值；
2、将左右两边较小的值减去当前值则为差值；
代码：

class Solution {public:    int trap(vector<int>& height) {        //这个方法算是一种常见的技巧，从两边各扫描一次得到我们需要维护的变量，通常适用于当前元素需要两边元素来                 //决定的问题，非常类似的题目是Candy，有兴趣的朋友可以看看哈。        //也可以直接找到最大的那个点，然后左一半分析，右一半分析        vector<int> leftmax(height.size());        //int m=height[0];        int maxleft=0;        for(int i=0;i<height.size();i++)        {            leftmax[i]=maxleft;            maxleft=maxleft>height[i]?maxleft:height[i];        }        vector<int> rightmax(height.size());        int maxright=0;        for(int i=height.size()-1;i>=0;i--)        {            rightmax[i]=maxright;            maxright=maxright>height[i]?maxright:height[i];        }        int water=0;        for(int i=0;i<height.size();i++)        {            int dep=min(leftmax[i],rightmax[i])-height[i];            if(dep>0) water+=dep;        }        return water;    }};