LeetCode[337] House Robber III

Dr_Noooo 发布于2019-08-14 15:00 / 778人阅读

摘要：复杂度思路对于每一个位置来说，考虑两种情况分别对和再进行计算。用对已经计算过的进行保留，避免重复计算。

LeetCode[337] House Robber III

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:

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

Example 2:

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

recursion + Memorization

复杂度
O(N), O(lgN)

思路
对于每一个位置来说，考虑两种情况, Max(child, subchild + root.val).
分别对child和subchild再进行recursion计算。

用map对已经计算过的node进行保留，避免重复计算。

代码

public int rob(TreeNode root) {
    // Base case;
    if(root == null) return 0;
    if(root.left == null && root.right == null) return root.val;
    if(map.containsKey(root)) return map.get(root);
    int child = 0, subchild = 0;
    if(root.left != null) {
        child += rob(root.left);
        subchild += rob(root.left.left) + rob(root.left.right);
    }
    if(root.right != null) {
        child += rob(root.right);
        subchild += rob(root.right.left) + rob(root.right.right);
    }
    int val = Math.max(child, subchild + root.val);
    map.put(root, val);
    return val;
}

海外云主机 GPU服务器 Robber House ee337新域名 III

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