House Robber II:打家劫舍 取非相邻元素求和最大，且认为第一个元素与最后一个元素相邻

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.

打家劫舍 取非相邻元素求和最大，且认为第一个元素与最后一个元素相邻

举例：1 1 1 输出1

1 2 3 1 输出4

0 输出0

1 3 输出3

思路：标准的动态规划，和普通的House Robber.类似，因为第一个元素与最后一个元素不能同时取，所以分别对0-（n-2）与1-（n-1）各位元素取符合题意的最值，再取最大者。

代码写的很啰嗦，可以看Leetcode上的简洁版┗|｀O′|┛ 嗷~~

class Solution {    int result = 0;      public int get(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 d[] = new int[nums.length+1];        d[0] = 0;        d[1] = nums[0];        d[2] = Math.max(nums[0],nums[1]);        for(int i = 3;i<=nums.length;i++){            d[i] = Math.max(d[i-1],d[i-2]+nums[i-1]);        }        return d[nums.length];    }               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[] a1 = new int[nums.length];        int[] a2 = new int[nums.length];        for(int i = 0;i<nums.length-1;i++){            a1[i] = nums[i];            a2[i] = nums[i+1];        }        return Math.max(get(a1),get(a2));                }}

简化版本：

private int rob(int[] num, int lo, int hi) {    int include = 0, exclude = 0;//对于数组中的每一个元素，都包含两种选法：选取当前元素include，不选取当前元素exclude    for (int j = lo; j <= hi; j++) {        int i = include, e = exclude;//include表示选取当前元素的选法，exclude表示不选取当前元素的选法        include = e + num[j];//如果选取当前元素，那么就得选择没有选取前一元素的选法，因为相邻元素不能同时取        exclude = Math.max(e, i);//对于不选取当前元素的选法，应该是之前一步中总和最大的两种选法    }    return Math.max(include, exclude);/选择两种选法中最大的选法}public int rob(int[] nums) {//由于第一个元素与最后一个元素不能同时选取，所以把他们分开成两组，在分别计算最优值，取最大者    if (nums.length == 1) return nums[0];    return Math.max(rob(nums, 0, nums.length - 2), rob(nums, 1, nums.length - 1));}