[leetcode]Trapping Rain Water

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! 

算法思路：

短板效应，蓄水量与桶的短板相关。求出height数组中的最长板，然后从两端向长板迭代，遇到更短的则求出蓄水量，遇到较长的，则更新边缘。

代码如下：

public class Solution {
    public int trap(int[] height) {
        if(height == null || height.length < 3) return 0;
        int max = 0;
        for(int i = 0; i < height.length;i++){
            if(height[i] > height[max]) max = i;
        }
        int high = 0,res = 0;
        for(int i = 0; i < max; i++){
            if(height[i] < height[high]){
                res += height[high] - height[i];
            }else if(height[i] > height[high]){
                high = i;
            }
        }
        high = height.length - 1;
        for(int i = height.length - 1; i > max; i--){
            if(height[i] < height[high]){
                res += height[high] - height[i];
            }else if(height[i] > height[high]){
                high = i;
            }
        }
        return res;
    }
}

具体算法分析可以参考这里。