首页 > 其他 > 详细

198. House Robber

时间:2016-09-15 13:41:46      阅读:191      评论:0      收藏:0      [点我收藏+]

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonightwithout alerting the police.

非常经典的小偷问题。怎么偷:如果偷现在的这家,那么总钱数就是现在的加上之前之前的。如果不偷,那么总钱数就是之前这家。比较两者哪个大来决定偷或者不偷。

public class Solution {
    public int rob(int[] nums) {
        if(nums.length==0)
        {
            return 0;
        }
        int[] dp=new int[nums.length+1];
        dp[0]=0;
        dp[1]=nums[0];
        for(int i=2;i<=nums.length;i++)
        {
            dp[i]=Math.max(dp[i-2]+nums[i-1],dp[i-1]);
        }
        return dp[nums.length];
    }
}

 

198. House Robber

原文:http://www.cnblogs.com/Machelsky/p/5874633.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!