[LeetCode]House Robber II

Question

Note: This is an extension of House Robber.

After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

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本题难度Medium。

题意

房子围城圈，抢头就不能抢尾。

DP

复杂度

时间 O(N) 空间 O(1)

思路

这道题难点在于围成圈，解决关键就在于最后那个房子的判断，实际上还是换汤不换药。设到达最后那个房子状态为f[N]，它只有两个可能：

1. 抢最后那个房子，那么就不能抢第一个房子
2. 不抢最后那个房子，那么f[N]=f[N-1]

我们只要另外用一个数组n[i]来表示不抢第一个房子的状态，结果就是max(f[N-1],n[N])。这里同样可以使用[LeetCode]House Robber 的优化，使用2对变量即可。

代码

public class Solution {    public int rob(int[] nums) {       //require       int N=nums.length;       if(N<2)return N==0?0:nums[0];       //a1、b1是正常的，a2、b2是不抢第一个房子       int a1=nums[0],b1=Math.max(a1,nums[1]),a2=0,b2=nums[1];       //invariant       for(int i=2;i<N;i++){           int tmp1=b1,tmp2=b2;           b1=Math.max(a1+nums[i],b1);           b2=Math.max(a2+nums[i],b2);           a1=tmp1;a2=tmp2;       }       //ensure       return Math.max(a1,b2); //注意不是max(b1,b2)          }}