[LeetCode] House Robber II

This problem is a little tricky at first glance. However, if you have finished the House Robber problem, 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

  1. Rob houses 0 to n - 2;
  2. Rob houses 1 to n - 1.

The code is as follows (some edge cases are handled explicitly).

 1     int robHelper(vector<int>& nums, int left, int right){
 2         if (left == right) return nums[left];
 3         vector<int> money(right - left + 1, 0);
 4         money[0] = nums[left];
 5         money[1] = max(nums[left], nums[left + 1]);
 6         for (int i = 2; i <= right - left; i++)
 7             money[i] = max(money[i - 1], money[i - 2] + nums[left + i]);
 8         return money[money.size() - 1];
 9     }
10     int rob(vector<int>& nums) {
11         if (nums.empty()) return 0;
12         int n = nums.size();
13         if (n == 1) return nums[0];
14         return max(robHelper(nums, 0, n - 2), robHelper(nums, 1, n - 1));
15     }

 

posted @ 2015-06-02 23:17  jianchao-li  阅读(172)  评论(0)  编辑  收藏  举报