[Leetcode] 213. House Robber II 解题报告

题目：

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：

如果窃取了第0号房子，那么就不能再窃取第n-1号房子；而如果不窃取第0号房子，那么就可以窃取第n-1号房子。所以我们可以将窃取方案按照是否窃取第0号房子分为两种类型，而每种类型的解法和House Robber的完全相同，我们最后取获利最大者即可。

假设dp[i]代表截止第i个房子可以获得的最大利润，那么递推公式为：dp[i] = max(dp[i-1], dp[i -  2] + nums[i])。可以看出dp[i]仅仅与dp[i-1]和dp[i-2]相关，所以我们可以进一步将空间复杂度从O(n)降低到O(1)。具体实现见下面的代码片段。

代码：

class Solution {public:    int rob(vector<int>& nums) {        int size = nums.size();        if(size == 0) {            return 0;        }        else if(size == 1) {            return nums[0];        }        return max(robBetween(nums, 0, size - 2), robBetween(nums, 1, size - 1));    }private:    int robBetween(vector<int>& nums, int start, int end) {        if(start == end) {            return nums[start];        }         int value0 = nums[start];        int value1 = max(value0, nums[start + 1]);        for(int i = start + 2; i <= end; ++i) {            int max_value = max(value0 + nums[i], value1);            value0 = value1;            value1 = max_value;        }        return value1;    }};