House Robber 非负数组，相邻不能相加，求最大的和是多少(动态规划)

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 andit 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.

出处：http://blog.csdn.net/xudli/article/details/44795737

这一篇更不错：http://www.programcreek.com/2014/03/leetcode-house-robber-java/

public class Solution {

    public int rob(int[] nums) {

        int n=nums.length;

        if(nums==null || n==0)  return 0;

        int[] n1=new int
;

        int[] n2=new int
;

        n1[0]=nums[0];

        n2[0]=0;

        for(int i=1;i<n;i++){

            n1[i]=n2[i-1]+nums[i];

            n2[i]=Math.max(n1[i-1],n2[i-1]);

        }

        return Math.max(n1[n-1],n2[n-1]);

    }

}

动态规划（Dynamic Programming）

状态转移方程：dp[i] = max(dp[i - 1], dp[i - 2] + num[i - 1])//多么的简洁明了啊

其中，dp[i]表示打劫到第i间房屋时累计取得的金钱最大值。

时间复杂度O(n)，空间复杂度O(n)

public class Solution {

    public int rob(int[] nums) {

        int n=nums.length;

        if(nums==null || n==0)  return 0;

        int[] dp=new int[n+1];//定义的长度比数组长1

        dp[0]=0;

        dp[1]=nums[0];

        for(int i=2;i<n+1;i++){

            dp[i]=Math.max(dp[i-1],dp[i-2]+nums[i-1]);

        }

        return dp
;

    }

}

这个方法对于我来说，比较容易理解

public class Solution {

    public int rob(int[] nums) {

        int n=nums.length;

        if(nums==null || n==0)  return 0;

        int odd=0;

        int even=0;

        for(int i=0;i<n;i++){

            if(i%2==0){

                even+=nums[i];

                even=even>odd?even:odd;

            }else{

                odd+=nums[i];

                odd=odd>even?odd:even;

            }

        }

        return Math.max(odd,even);

    }

}