House Robber II

https://leetcode.com/problems/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.

解题思路：

前面一题的follow-up，多了一个限制条件，第一个和最后一个房子，也算相邻，所以只能偷一个。这样在计算到dp[nums.length]的时候，不能再简单的max(dp[i - 2] + nums[i], dp[i - 1])了。前面的dp[i - 2]不能包含第一个房子。问题是，你不知道这个dp[i-2]是不是偷了第一个房子，因为子状态的定义是，偷到第i个房子时候的最大值。

所以，这题可以分解为两个问题，1)偷第一个房子，时候的最大值，2)偷最后一个房子，时候的最大值。

dp两次，再求最大值。

public class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0) {
            return 0;
        }
        if(nums.length == 1) {
            return nums[0];
        }
        int prepre = 0;
        int pre = 0;
        int result = pre;
        int max1 = 0;
        for(int i = 0; i < nums.length - 1; i++) {
            result = Math.max(pre, prepre + nums[i]);
            prepre = pre;
            pre = result;
        }
        max1 = result;

        int max2 = 0;
        prepre = 0;
        pre = 0;
        result = pre;

        for(int i = 1; i < nums.length; i++) {
            result = Math.max(pre, prepre + nums[i]);
            prepre = pre;
            pre = result;
        }
        max2 = result;
        return Math.max(max1, max2);
    }
}