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！

注意对每根柱子能盛的水并不是自身能决定的，而是取决于其左右两边的柱子。

先记录最高的柱子maxHeight，把数组一分为二，分别计算最高柱子左右两边的盛水量。

对于最高柱子左边的部分，

　　从首端开始扫描，并使用leftHeight记录已扫描部分的中得最高柱子。

　　如果leftHeight > curHeight，说明当前柱子能盛水，当前柱子所盛水的容量为leftHeight-curHeight

　　如果leftHeight<= curHeight，说明当前柱子不能盛水，则更新leftHeight = curHeight

对于最高柱子右边的部分，从末端开始倒序扫描，求右半部分的水，即可

class Solution {

public:

    int trap(int A[], int n) {

        int maxHeightIndex = ;

        for(int i =  ; i < n; ++ i){

            if(A[i] > A[maxHeightIndex]) maxHeightIndex = i;

        }

        int leftHeight = ,res = ;

        for(int i =  ; i < maxHeightIndex;++ i){

            if(leftHeight > A[i]) res+=leftHeight-A[i];

            else leftHeight = A[i];

        }

        int rightHeight = ;

        for(int i = n-; i>maxHeightIndex; --i){

            if(rightHeight > A[i]) res+=rightHeight-A[i];

            else rightHeight = A[i];

        }

        return res;

    }

};