Leetcode 213. House Robber II-优快云博客

本文链接：https://blog.youkuaiyun.com/m0_46229882/article/details/116104942

本文介绍了一个编程问题，涉及如何在不触动警报系统的情况下，最大化在一个圆形房屋区域内的盗窃金额。通过分析给定金钱分布，设计了一种动态规划方法来确定最佳的偷窃路径，以避免相邻房屋触发警报。

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You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
Example 2:

Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.
Example 3:

Input: nums = [0]
Output: 0

考虑到，最后一个和第一个房子是不能够同时进入的
所以分为两种情况
0.不包括最后一个屋子~就抢劫0~n-2号屋子
1.不包括第一个屋子~就抢劫1~n-1号屋子
这样的话，return上面两种情况的最大值就好了
*/
class Solution {
public:
    int rob(vector<int>& nums) {
        int n=nums.size();
        int f[n+10];
        if(n==1)return nums[0];
        else if(n==2)return max(nums[0],nums[1]);
        return max(func(nums, 0, n-2),func(nums, 1, n-1));  //
    }
    int func(vector<int>& nums, int begin, int end) {
        int n = end - begin + 1;  //
        int f[n+5];
        f[1]=nums[begin]; //
        f[2]=max(nums[begin],nums[begin+1]); //
        for(int i=3;i<=n;i++){
            f[i]=max(f[i-1],f[i-2]+nums[begin+i-1]);//nums[i-1]也随着begin(0 or 1)的不同而不同
        }
        return f[n];
    }
};
解题思想参考柳神的，附上链接(https://www.liuchuo.net/archives/1177)