LeetCode算法题目：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.

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分析：

• 首先找到最高的柱子，将数组分为两半。
• 从左往右扫描，每次找到当前的最大值和其右侧比较，若比他大，替换掉他，否则用他减去这个柱子的值得到此柱子的蓄水量。
• 从右往左扫描，做同样处理。

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代码实现:

class Solution {public:    int trap(vector<int>& height) {        int n = height.size();        int max=0;        for(int i=0;i<n;i++){            if(height[i]>height[max])               max=i;        }        int  sum=0;        for(int i=0,left=0;i<max;i++){            if(height[i]>left)                 left=height[i];            else                 sum=sum+left-height[i];        }        for(int j=n-1,right=0;j>max;j--){            if(height[j]>right)                 right=height[j];            else                 sum=sum+right-height[j];        }        return sum;    }};