Home Robber II

题目描述

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.

思路分析

这题目是Home Robber【https://leetcode.com/problems/house-robber/description/】的进阶版。所有的houses围成一个圆圈，robber不能同时偷窃第1间house和最后1间house，因此将问题分为：不选择第1间house，或者不选择最后1间house，最后再取最大值。
子函数rob(vector& nums, int start, int end)计算出从第start个house到第end个house偷窃的钱的最大值，假设dp[i]表示从第start个house到第i个house偷窃的最大值，则针对robber不选择第i个house和选择第i个house两种情况，此动态规划问题的状态转移方程：dp[i]=max(dp[i-1],dp[i-2]+nums[i])。

代码实现

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