House Robber II

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.

It’s a very similar question to House Robber I, except the houses are circled. The first house and the last house are considered as adjacent.

Find a start point, say nums[0]. Consider two situations, rob it or not. Discuss them separately, and find the maximum result.

//Just decide to rob the first room or not, 
//divide it into two situations.

class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.size() == 0) return 0;
        if(nums.size() == 1) return nums[0];
        if(nums.size() == 2) return max(nums[0], nums[1]);
        //n == 3?
        
        //case 1 rob nums[0]
        int dp[nums.size()];
        memset(dp, 0, sizeof(dp));
        dp[0] = nums[0];
        dp[1] = 0;
        for(size_t  i = 2; i < nums.size(); i++){
            for(size_t j = 0; j < i - 1; j++){
                int value = nums[i] + dp[j];
                if(dp[i] < value){
                    dp[i] = value;
                }
            } 
        }
        int ans1 = max(dp[nums.size() - 2], dp[nums.size() - 3]); // do not rob the last house
        memset(dp, 0, sizeof(dp));
        dp[0] = 0;
        dp[1] = nums[1];
        for(size_t  i = 2; i < nums.size(); i++){
            for(size_t j = 0; j < i - 1; j++){
                int value = nums[i] + dp[j];
                if(dp[i] < value){
                    dp[i] = value;
                }
            } 
        }
        int ans2 = max(dp[nums.size() - 1], dp[nums.size() - 2]);
        return max(ans1, ans2);
    }
};

Again, walk through is very important.