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!

解法：

对数组，至少有一个最高点，

1. 两头向中间扫描，左边找到局部最高点leftmaxLocal，当当前高度比leftmaxLocal小，则area+=(leftmaxLocal-curheight),否则更新leftmaxLocal；同理右边也是这样。

   
   

   
   

int trap(vector<int>& height) {

        int size=height.size();

        if(size<3)

            return 0;

        

        int left=0,right=size-1;

        int lhm=0,rhm=0;

        int lhc=0,rhc=0;

        int area=0;

        while(left<=right){

            lhc=height[left];

            rhc=height[right];

            if(lhc<rhc){

                if(lhm>lhc)

                    area+=(lhm-lhc);

                else

                    lhm=lhc;

                left++;

            }

            else{

                if(rhm>height[right]){

                    area+=(rhm-height[right]);

                }

                else

                    rhm=height[right];

                right--;

            }

        }

        return area;

    }