Description Link to heading

213.house-robber-ii

Solution Link to heading

Referring to 198.house-robber, there be another constraint that first and last can’t be selected at the same time. So we can split the array into two part: one for [0, n - 1), another for [1, n), corresponding to dp0 and dp1 respectively, just return max(dp0, dp1).

Code Link to heading 

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 0) return 0;
        if (nums.size() == 1) return nums[0];
        int result1 = robRange(nums, 0, nums.size() - 2); 
        int result2 = robRange(nums, 1, nums.size() - 1); 
        return max(result1, result2);
    }

    int robRange(vector<int>& nums, int start, int end) {
        if (end == start) return nums[start];
        vector<int> dp(nums.size());
        dp[start] = nums[start];
        dp[start + 1] = max(nums[start], nums[start + 1]);
        for (int i = start + 2; i <= end; i++) {
            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
        }
        return dp[end];
    }
};