leetcode_House Robber II

描述：

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.

思路：

1.和House Robber的思路类似，只是本题中房子是首尾相连的，也就是说第一个房子和最后一个房子不能都被盗，否则，触发警报，小偷被抓

2.在House Robber的基础上，a:如果从第一座房子开始偷窃，那么最后一个房子肯定不能被偷；b:如果从第二座房子偷窃，那么最后一个房子可以被偷，分别对这两种情况进行计算即可。

代码：

 public int rob(int[] nums) {
        if(nums==null)
            return 0;
        int len=nums.length;
        int maxProfit=0;
        if(len==0)
            return 0;
        else if(len==1)
            return nums[0];
        else if(len==2)
            return Math.max(nums[0],nums[1]);
        else
        {
            int tempMaxProfit1=getMaxProfit(nums, 0, nums.length-2);
            int tempMaxProfit2=getMaxProfit(nums, 1, nums.length-1);
            maxProfit=Math.max(tempMaxProfit1, tempMaxProfit2);
        }
        return maxProfit;
    }
	public int getMaxProfit(int nums[],int start,int end)
	{
		int maxProfit1=nums[start];//用maxProfit1、maxProfit2来避免用数组来存储临时变量
        int maxProfit2=Math.max(nums[start],nums[start+1]);
        int maxProfit=Math.max(maxProfit1, maxProfit2);
        for(int i=start+2;i<=end;i++)
        {
            maxProfit=Math.max(maxProfit2,maxProfit1+nums[i]);
            maxProfit1=maxProfit2;
            maxProfit2=maxProfit;
        }
        return maxProfit;
	}