LeetCode 213. House Robber II（小偷游戏）

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本文介绍了一种解决LeetCode上环形房屋抢劫者II问题的方法。该问题要求在不触动警报的情况下，计算出从一系列排成环形的房屋中能抢到的最大金额。通过动态规划将问题分解为两部分进行求解。

原题网址：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.

方法：动态规划，根据是否从第一个房子开始，分开两种情况计算。

public class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        if (nums.length == 2) return Math.max(nums[0], nums[1]);
        int max = 0;
        int[] money = new int[nums.length];
        money[0] = nums[0];
        money[1] = Math.max(nums[0], nums[1]);
        for(int i=2; i<nums.length-1; i++) {
            money[i] = Math.max(money[i-2] + nums[i], money[i-1]);
        }
        max = Math.max(max, money[nums.length-2]);
        Arrays.fill(money, 0);
        money[1] = nums[1];
        money[2] = Math.max(nums[1], nums[2]);
        for(int i=3; i<nums.length; i++) {
            money[i] = Math.max(money[i-2] + nums[i], money[i-1]);
        }
        max = Math.max(max, money[nums.length-1]);
        return max;
    }
}