LeetCode-House Robber II

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.

Analysis:

First method: consider two cases (a) rob first house, (b) not rob first house. For each case, we calculate two arrays robThis[i] and notRobThis[i], i.e., the max profit we can get with/without robbing ith house, and we always have

robThis[i] = notRobThis[i-1]+nums[i];

notRobThis[i] = max(robThis[i-1], notRobThis[i-1]);

At last, we just compare (a) rob first house case's notRobThis[n], (b) not rob first house case's robThis[n]. As we are not allowed to rob both of 1st and last houses.

Solution:

public class Solution {
    public int rob(int[] nums) {
        if (nums.length==0) return 0;
        if (nums.length==1) return nums[0];

        int rFirstAThis = nums[0];
        int rFirstNoThis = 0;
        int nrFirstRThis = 0;
        int nrFirstNoThis = 0;

        for (int i=1;i<nums.length;i++){
            int pre = rFirstAThis;
            rFirstAThis = rFirstNoThis + nums[i];
            rFirstNoThis = Math.max(pre,rFirstNoThis);

            pre = nrFirstRThis;
            nrFirstRThis = nrFirstNoThis + nums[i];
            nrFirstNoThis = Math.max(pre,nrFirstNoThis);
        }

        return Math.max(rFirstNoThis,nrFirstRThis);
    }
}

Solution 2:

We can just exclude 1st house / last house and do house robber I twice, and compare the results.

Actually the same idea, but different implementation.