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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

思路：House Robber的变形，因为牵扯到圆环问题，所以首尾不能相连，
需要遍历两次：0-nums.length-2和1-nums.length-1，然后返回一个最大值。

public class Solution {
    public int rob(int[] nums) {
        if(nums.length==0){
            return 0;
        }
        if(nums.length==1){
            return nums[0];
        }
        if(nums.length==2){
            return Math.max(nums[0],nums[1]);
        }
        int m1=rob(nums,0,nums.length-2);
        int m2=rob(nums,1,nums.length-1);
        return Math.max(m1,m2);
    }
    private int rob(int[]nums, int s, int e){
        int[] sum = new int[e-s+1];
        sum[0]=nums[s];
        sum[1]=Math.max(nums[s],nums[s+1]);//important
        for(int i=s+2;i<=e;i++){
            sum[i-s]=Math.max(sum[i-1-s],sum[i-2-s]+nums[i]);//nums[i]important
        }
        return sum[e-s];
    }
}