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 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.

思路

題目大意

求一條環路上的所求最大和，求和時，環路兩個相鄰的元素不能同時計算。

解題思路

基於 198. House Robber 問題 的變型。 總體解法還是 House Robber 的dp。最後結果為 nums[0:nums.length-2] 與 nums[1:nums.length-1] 的House Robber 解 的 較大值。 這是因為 nums[0] 和 nums[nums.length-1] 同時只有一個可以加和。

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 1)    return nums[0];
        return Math.max(eachRob(nums,0,nums.length-2),eachRob(nums,1,nums.length-1));
    }
    public int eachRob(int[] nums,int first,int last){
        int a = 0,b = 0;
        for(int i = first; i<=last;i++){
            int c = Math.max(a+nums[i],b);
            a = b;
            b = c;
        }
        return b;
    }     
}