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;
}