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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], return6.
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：
1 public class Solution { 2 public int trap(int[] height) { 3 if(height == null || height.length < 3) return 0; 4 int max = 0; 5 for(int i = 0; i < height.length;i++){ 6 if(height[i] > height[max]) max = i; 7 } 8 int high = 0,res = 0; 9 for(int i = 0; i < max; i++){ 10 if(height[i] < height[high]){ 11 res += height[high] - height[i]; 12 }else if(height[i] > height[high]){ 13 high = i; 14 } 15 } 16 high = height.length - 1; 17 for(int i = height.length - 1; i > max; i--){ 18 if(height[i] < height[high]){ 19 res += height[high] - height[i]; 20 }else if(height[i] > height[high]){ 21 high = i; 22 } 23 } 24 return res; 25 } 26 }
具体算法分析可以参考这里。
[leetcode]Trapping Rain Water,布布扣,bubuko.com
标签：style blog http color os io strong for
原文地址：http://www.cnblogs.com/huntfor/p/3898901.html
