# An easy-understanding solution in C++

• This problem is an advanced version of Problem "House Robber", the only difference is that the houses are arranged in a circle, which means the first house is adjacent to the last house.

From a global view, any rob solution has two possible cases: rob the first house, or not.

If we rob the first house, we can't rob the last house, so the problem transfer to "how to rob in house[1, n-1]". If we do not rob the first house, the problem transfer to "how to rob in house[2, n]".

Assuming that we have understand the solution for problem "House Robber", now we can simply design the new strategy as below:

``````class Solution {
public:
int rob(vector<int> &nums) {
if (nums.size() == 0) {
return 0;
}
if (nums.size() == 1) {
return nums.at(0);
}
vector<int> case1(nums);
vector<int> case2(nums);

vector<int>::iterator v1 = case1.begin();
case1.erase(v1);
case2.pop_back();

int maxRobValue1 = simpleRob(case1);
int maxRobValue2 = simpleRob(case2);
int maxRobValue = max(maxRobValue1, maxRobValue2);
return maxRobValue;
}
int simpleRob(vector<int> &num){
int *f = new int[num.size() + 1];
f[0] = 0;
for (int i = 1; i <= num.size(); i++) {
if (i == 1) {
f[i] = num.at(i-1);
}
else {
f[i] = max(f[i-2] + num.at(i-1), f[i-1]);
}
}
int robMaxValue = f[num.size()];
return robMaxValue;
}
};``````

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.