[LeetCode] House Robber II

This problem is a little tricky at first glance. However, if you have finished the House Robberproblem, this problem can simply be decomposed into two House Robber problems. Suppose there are

n

houses, since house

0

and

n - 1

are now neighbors, we cannot rob them together and thus the solution is now the maximum of

Rob houses

0

to

n - 2

;

Rob houses

1

to

n - 1

.

The code is as follows. Some edge cases (

n <= 2

) are handled explicitly.

class Solution {
public:
int rob(vector<int>& nums) {
int n = nums.size();
if (n <= 2) return n ? (n == 1 ? nums[0] : max(nums[0], nums[1])) : 0;
return max(robber(nums, 0, n - 2), robber(nums, 1, n -1));
}
private:
int robber(vector<int>& nums, int l, int r) {
int pre = 0, cur = 0;
for (int i = l; i <= r; i++) {
int temp = max(pre + nums[i], cur);
pre = cur;
cur = temp;
}
return cur;
}
};