Trapping Rain Water 左右指针寻找最大容量的水

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!

class Solution {public://首先找到最高点，然后从最左边开始遍历，保存一个次高点e，然后遍历，只要当前点小于这个e，水量加，否则更新次高点//右边同理    int trap(vector<int>& height) {                int len=height.size();        if(len<1)            return 0;                    int maxID=0;        for(int i=1;i<len;i++)        {            if(height[i]>height[maxID])                maxID=i;        }                int sum=0;        int curHeight=height[0];        for(int i=1;i<maxID;i++)        {            if(height[i]>curHeight)                curHeight=height[i];            else                sum+=curHeight-height[i];        }                curHeight=height[len-1];        for(int i=len-2;i>maxID;i--)        {            if(height[i]>curHeight)                curHeight=height[i];            else                sum+=curHeight-height[i];        }                return sum;    }};