题解：Trapping Rain Water

题目如下：

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!

解题方面，从前到后遍历，判断左右高度后累增水的数量，这个数量应该要是较低一边的数量（较高的减去较小的），并且坐标有相应的增大/减小。设置的两个坐标一个从最左边开始一个从最右边开始，依据前面的规则向中间靠拢。当左边坐标不再小于右边完成。代码如下：

int trap(vector<int>& height) {int result = 0;    int i = 0, k = height.size() - 1;    int h;    while (i < k) {        if (height[i] < height[k]) {            h = height[i++];            while (i < k && height[i] <= h) {                result += h - height[i];                i++;            }        }else {            h = height[k--];            while (i < k && height[k] <= h) {                result += h - height[k];                k--;            }        }    }    return result;}