[leetCode] 213. House Robber II Java

题目：

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.

题意及分析：这里相比 House Robber 多了一个约束条件，即数组第一个和最后一个是相邻的。所以第一个和最后一个不能同时选，那么我们可以遍历两次：

第一遍计算[0,n-2]，也就是选择第一个房子到倒数第二个房子，求最大值

第二遍计算[1,n-1]，也就是从第二个房子到最后一个房子，求最大值

两个最大值比较的较大值就是结果值。

代码：

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]);  
        return Math.max(robsub(nums, 0, nums.length-2), robsub(nums, 1, nums.length-1));  
    }  
      
    private int robsub(int[] nums, int s, int e) {  
        int n = e - s + 1;  
        int[] d =new int[n];  
        d[0] = nums[s];  
        d[1] = Math.max(nums[s], nums[s+1]);  
          
        for(int i=2; i<n; i++) {  
            d[i] = Math.max(d[i-2]+nums[s+i], d[i-1]);  
        }  
        return d[n-1];  
    }  
}