LeetCode House Robber2


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.

The original problem is here.

The original code is here.

My Solution

I solve this problem in C++, as below:

*House Robber
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
using namespace std;

class Solution {
    int rob(vector<int>& nums) {
        int size = nums.size();
        if(size <= 0)
            return 0;
        if(size == 1)
            return nums[0];        
        if(size == 2)
            return max(nums[0], nums[1]);
        vector<int> money = vector<int>(size, 0);
        vector<int> index = vector<int>(size, 0);
        //Dynamic programming
        money[0] = nums[0];
        index[0] = 1;
        money[1] = max(nums[0],nums[1]);
        if(nums[0] > nums[1])
            index[1] = 1;
            index[1] = 0;
        for(int count=2;count<size;++count){
            money[count] = max(money[count-1],money[count-2]+nums[count]);            
            if( money[count-1] >= money[count-2]+nums[count] ){
                if(count == size-1)
                    return money[count];
                index[count] = index[count-1];
                index[count] = index[count-2];
        if(index[size-1] == 0)
            return money[size-1];
        //The result without the last one 
        int max1 = money[size-2];
        money[1] = nums[1];
        money[2] = max(nums[1],nums[2]);
        //Compute the result without the first one
        for(int count=3;count<size;++count){
            money[count] = max(money[count-1],money[count-2]+nums[count]);            
        //return the max of the two results
        return max(money[size-1], max1);


To solve the problem, dynamic programming is used. Assume the robber get one house i, he just need to compute the maximum of the nums[i-1] and (nums[i-2]+nums[i]) as the money[i], this will not alert police. In the end, the robber can get the maximum of money[end-1]; Additionally, the first and the end of the house can’t be robbed at the same time.