leetcode No42. Trapping Rain Water

Question:

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!

数组元素代表高度，如图所示，求能装多少水

Algorithm：

分别从左右两边向最高点逼近，只在元素小的一边地方储水。由左边向最高点逼近时，设左边当前的最大值leftmax,当前遍历到i，如果leftmax > A[i]; sum += (leftmax- A[i]);否则，currentMax = i，此条带无法储水；由右边逼近最高点类似左边。

Accepted Code:

class Solution {
public:
    int trap(vector<int>& height) {
        int left=0;   
        int right=height.size()-1;
        int leftmax=0;   //左边最大值
        int rightmax=0;  //右边最大值
        int res=0;       //结果
        while(left<=right)
        {
            if(height[left]<=height[right])
            {
                if(height[left]>leftmax)
                    leftmax=height[left];
                else
                    res+=leftmax-height[left];
                left++;
            }
            else
            {
                if(height[right]>rightmax)
                    rightmax=height[right];
                else
                    res+=rightmax-height[right];
                right--;
            }
        }
        return res;
    }
};