題目

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 a 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 nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:
Input: nums = [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: nums = [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.

Example 3:
Input: nums = [0]
Output: 0

Constraints:
1 <= nums.length <= 100
0 <= nums[i] <= 1000

思路

因為nums[0]和nums[nums.length-1]不可兼得，所以選擇比較從0到n-1個數和1到n個數的子問題，選擇答案較大的那個為解。對於子問題採用簡單的dp解決，遞推式為$dp[i]=max(dp[i-1],dp[i-2]+nums[i])$。

程式碼

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        
        return Math.max(help(nums, 0, nums.length - 1), help(nums, 1, nums.length));
    }
    
    public int help(int[] nums, int start, int end) {
        int pprev = 0, prev = 0, curr = 0;
        for (int i = start; i < end; i++) {
            curr = Math.max(pprev + nums[i], prev);
            pprev = prev;
            prev = curr;
        }
        return curr;
    }
}