[Leetcode] 337. House Robber III | 打家劫舍 III

yPhantom 2020年02月05日 24次浏览

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

Example 1:

Input: [3,2,3,null,3,null,1]

     3
    / \
   2   3
    \   \ 
     3   1

Output: 7 
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

Input: [3,4,5,1,3,null,1]

     3
    / \
   4   5
  / \   \ 
 1   3   1

Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

这道题求小偷能抢的最大money,小偷不能抢直接连接的两个结点。Disscuss区有个讲的十分完美的帖子:

Step by step tackling of the problem

从最基础的递归解法一步步讲解到最优解法,逻辑十分清晰。

最基础的递归解法中,存在重复计算,因此在第二种解法中增加一个HashMap来避免重复计算。而在第三种解法中,直接从方案本身避免的重复计算产生。值得学习。

public int rob(TreeNode root) {
    int[] res = robSub(root);
    return Math.max(res[0], res[1]);
}

private int[] robSub(TreeNode root) {
    if (root == null) return new int[2];
    
    int[] left = robSub(root.left);
    int[] right = robSub(root.right);
    int[] res = new int[2];

    res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
    res[1] = root.val + left[0] + right[0];
    
    return res;
}

这道题表现出了动态规划问题的特征,最优子结构。对于每个结点都有被抢和不被抢两种情况,有点背包问题的味道。