House Robber II -- leetcode

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.

基本思路：

将此问题分解为两个子问题。

设屋子总数为n。

1. 抢第1间屋。此时，最后一间屋不能抢了。则可抢范围是[0, n-1]

2. 不抢第1间屋。此时，最后一间屋可以抢。 则可抢范围是[1, n]

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 1)
            return nums[0];
        return max(rob(nums, 0, nums.size()-1), rob(nums, 1, nums.size()));
    }
    
    int rob(vector<int>& nums, int start, int stop) {
        int last_last = 0, last = 0;
        for (int i=start; i<stop; i++) {
            int temp = max(last_last+nums[i], last);
            last_last = last;
            last = temp;
        }
        return last;
    }
};

House Robber II -- leetcode

原文：http://blog.csdn.net/elton_xiao/article/details/47702551