[LeetCode]213 强盗入屋 II

House Robber II(强盗入屋 II)

【难度：Medium】
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.
给定一个整数数组（首尾相连）表示一系列连续的屋子里面放有的金钱，在不能连续进入相邻屋子抢钱的情况下求能抢到的最大金额。

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解题思路

http://blog.youkuaiyun.com/qq_14821023/article/details/50880893
在198题的基础上，考虑到首尾相连的要求，将问题划分为在0~n-1和1~n两个区间动态规划，取两者中较大值。因为如果在0点抢钱的话，第n个点就不能抢；在1点抢的话，0点就不能抢。

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c++代码如下：

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.empty())
            return 0;
        int n = nums.size();
        if (n == 1)
            return nums[0];
        return max(rob_line(nums,0,n-1),rob_line(nums,1,n));
    }
    int rob_line(vector<int> nums, int start, int end) {
        int len = end-start+1;
        int rob[len];
        rob[0] = 0;
        rob[1] = nums[start];
        for (int i = 2; i < len; i++) {
            rob[i] = (rob[i-2]+nums[i+start-1])>rob[i-1] ?(rob[i-2]+nums[i+start-1]):rob[i-1];
        }
        return rob[len-1];
    }
};