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;
    }
};