198. House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

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.

Example 1:

Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

。第1或第2家吹大的開始，rob1+ n間隔rob2，之後就循環間隔。

Java

# T:O(n), S:O(n)
class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0) return 0;
        int temp = 0, rob1 = 0, rob2 = 0; 
        //swap
        for(int n: nums){
            temp = Math.max(rob1+n, rob2);
            rob1 = rob2;
            rob2 = temp;
        }
        return rob2;
    }
}

Python

# T:O(n), S:O(n)
class Solution:
    def rob(self, nums: List[int]) -> int:
        if nums == []:  return 0        
        rob1, rob2 = 0, 0
        # [rob1, rob2, n, n+1, ...]
        for n in nums:
            temp = max(rob1+ n, rob2)
            rob1 = rob2
            rob2 = temp
        return rob2

337.House Robber III

Tree structure

class Solution {
    public int rob(TreeNode root) {
        if(root == null)
            return 0;
        int[] result = helper(root);
        return Math.max(result[0], result[1]);
    }
    public int[] helper(TreeNode root){
        if(root == null) return new int[]{0,0};
        int[] left = helper(root.left);
        int[] right = helper (root.right);

        int rob = root.val+ left[1]+ right[1];
        int notRob = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        return new int[]{rob, notRob};
    }
}

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