213. House Robber II

Leetcode Dynamic Programming

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it 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 tonight without alerting the police.

Example 1:

Input: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

分析

这道题目是LeetCode 198. House Robber的延伸，既然已经知道了House Robber怎么写的，那么就可以把这道题目分解了。由于房子是圆形排列的，因此，不能同时抢首尾的房子，也就是说有两种抢法：0~last-1和1~last。

public int rob(int[] nums) {
    if (nums == null) return 0;
    int n = nums.length;
    if (n == 1) return nums[0];
    return Math.max(rob(nums, 1, n), rob(nums, 0, n - 1));
}

private int rob(int[] nums, int start, int end) {
    int last = 0, secondLast = 0, cur = 0;
    for (int i = start; i < end; i++) {
        cur = Math.max(nums[i] + secondLast, last);
        secondLast = last;
        last = cur;
    }
    return cur;   
}