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;
}