LeetCode(213) House Robber II

题意：和House Robber I一样，只是首尾两个数也不能同时取。

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.

解法：直接套用House Robber I的解法。既然A[0]和A[n-1]不能同时取，那么就算一下A[0]，A[1]，...，A[n-2]和A[1]，A[2]，...，A[n-1]这两个结果,取大的那个。复杂度：O（n）。
代码：
class Solution {
public:
int rob(vector<int>& nums) {
int n=nums.size();
if(n==1) return nums[0];
return max(myrob(nums,0,n-1),myrob(nums,1,n));
}
int myrob(vector<int>&nums,int begin,int end)
{
int n=end-begin;
if(n<=0) return 0;
vector<int> dp(n);
dp[0]=nums[begin];
dp[1]=max(nums[begin],nums[begin+1]);
for(int i=2;i<n;i++) dp[i]=max(dp[i-1],dp[i-2]+nums[begin+i]);
return dp[n-1];
}
};