213. 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.

这个题有环，因此0和最后一个，不能同时出现，怎么解决呢？计算1到n-1的值和2到n的值，取大的。。。

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 val1 = robHelper1(nums);        int val2 = robHelper2(nums);        return Math.max(val1, val2);    }        private int robHelper1(int[] nums) {  //0- n-2        int[] tmp = new int[nums.length];        tmp[0] = nums[0];        tmp[1] = nums[1];        for (int i = 2; i <= nums.length - 1; i++) {            tmp[i] = tmp[i - 2] + nums[i];            tmp[i - 1] = Math.max(tmp[i - 2], tmp[i - 1]);        }        return Math.max(tmp[nums.length - 2], tmp[nums.length - 3]);    }        private int robHelper2(int[] nums) {  //1- n-1        for (int i = 3; i <= nums.length - 1; i++) {            nums[i] = nums[i - 2] + nums[i];            nums[i - 1] = Math.max(nums[i - 2], nums[i - 1]);        }        return Math.max(nums[nums.length - 1], nums[nums.length - 2]);    }}