42. Trapping Rain Water

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!

本题采用动态规划的思想，时间复杂度O（n）；思路并不复杂，从左往右，扫描数组，找到一个大于等于left边界的值，计算局部sum值，再更新left边界。有一种特殊情况，就是右边没有值比左边边界高，这种情况通常出现在数组的最右边，或者是整个数组就是这样的结构，这种处理方法与上面的处理情况类似，反过来处理即可。

程序如下所示：

class Solution {    public int calLeftOrderSum(int[] height, int left, int right){        int sum = 0;        for (int i = left; i < right; ++ i){            sum += height[left] - height[i];        }        return sum;    }        public int calRightOrderSum(int[] height, int left, int right){        int sum = 0;        for (int i = right; i > left; -- i){            sum += height[right] - height[i];        }        return sum;    }        public int trap(int[] height) {int sum = 0, left = 0, len = height.length;        for (int i = 0; i < len; ++ i){            if (height[i] >= height[left]){                sum += calLeftOrderSum(height, left, i);                left = i;            }        }        int right = len - 1;        if (left < len){            for (int i = right; i >= left; -- i){                if (height[i] >= height[right]){                    sum += calRightOrderSum(height, i, right);                    right = i;                }            }        }        return sum;    }}