[Leetcode]House Robber II

House Robber II
Total Accepted: 18212 Total Submissions: 63419 Difficulty: Medium

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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

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遍历两次。
第一次假设不存在nums[len - 1],从1 ～ len - 2,得到一个最大值。
第二次假设不存在nums[0],从len - 2 ~ 1,得到一个最大值。
则真正的最优解肯定在这两个之中

class Solution {public:    int rob(vector<int>& nums) {        int len = nums.size();        if(!len)    return 0;        vector<int> get(len,0);        vector<int> pop(len,0);        get[0] = nums[0];        int res0 = nums[0];        for(int i = 1;i <= len - 2;++ i){            get[i] = max(pop[i - 1],i >= 2 ? get[i - 2] : 0) + nums[i];            pop[i] = max(pop[i - 1],get[i - 1]);            res0 = max(res0,max(get[i],pop[i]));        }        int res1 = nums[len - 1];        fill(get.begin(),get.end(),0);        fill(pop.begin(),pop.end(),0);        get[len - 1] = nums[len - 1];        for(int i = len - 2;i >= 1;-- i){            get[i] = max(pop[i + 1],i + 2 <= len - 1 ? get[i + 2] : 0) + nums[i];            pop[i] = max(pop[i + 1],get[i + 1]);            res1 = max(res1,max(get[i],pop[i]));        }        return max(res1,res0);    }};