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[LeetCode]Trapping Rain Water

时间:2014-09-23 16:46:54      阅读:224      评论:0      收藏:0      [点我收藏+]

标签:java   leetcode   

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.

bubuko.com,布布扣

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!

参考:戴方勤 (soulmachine@gmail.com)     https://github.com/soulmachine/leetcode

public class Solution {
    public int trap(int[] A) {
        int sum = 0;
    	int maxLeft [] = new int[A.length];
        int maxRight [] = new int [A.length];
        for(int i=1;i<A.length;i++){
        	maxLeft[i] = Math.max(maxLeft[i-1], A[i-1]);
        	maxRight[A.length-1-i] = Math.max(maxRight[A.length-i], A[A.length-i]);
        }
        for(int i=0;i<A.length;i++){
        	int height = Math.min(maxLeft[i],maxRight[i]);
        	if(height>=A[i]){
        		sum += height-A[i];
        	}
        }
        return sum;
    }
}





[LeetCode]Trapping Rain Water

标签:java   leetcode   

原文地址:http://blog.csdn.net/guorudi/article/details/39498087

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