leetcode 2368. Reachable Nodes With Restrictions（python）

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描述

There is an undirected tree with n nodes labeled from 0 to n - 1 and n - 1 edges. You are given a 2D integer array edges of length n - 1 where edges[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree. You are also given an integer array restricted which represents restricted nodes.

Return the maximum number of nodes you can reach from node 0 without visiting a restricted node. Note that node 0 will not be a restricted node.

Example 1:

Input: n = 7, edges = [[0,1],[1,2],[3,1],[4,0],[0,5],[5,6]], restricted = [4,5]
Output: 4
Explanation: The diagram above shows the tree.
We have that [0,1,2,3] are the only nodes that can be reached from node 0 without visiting a restricted node.

Example 2:

Input: n = 7, edges = [[0,1],[0,2],[0,5],[0,4],[3,2],[6,5]], restricted = [4,2,1]
Output: 3
Explanation: The diagram above shows the tree.
We have that [0,5,6] are the only nodes that can be reached from node 0 without visiting a restricted node.

Note:

2 <= n <= 10^5
edges.length == n - 1
edges[i].length == 2
0 <= a_i, b_i < n
a_i!= b_i
edges represents a valid tree.
1 <= restricted.length < n
1 <= restricted[i] < n
All the values of restricted are unique.

解析

根据题意，有一棵无向树，其 n 个节点标记为从 0 到 n - 1 和并且有 n - 1 条边。给定一个长度为 n - 1 的二维整数数组 edges ，其中 edges[i] = [a_i, b_i] 表示在树中有一条节点 a< sub>i 到 b_i 的边。给定一个整数数组 restricted ，它表示受限节点。返回可以从节点 0 开始可以到达的所有节点数，而无需访问受限节点。请注意，节点 0 不受限节。

其实这道题考察的就是 BFS ，我们首先要遍历所有的 edges ，将节点之间的边保存到字典 d 中，然后我们定义一个集合来保存可以访问的节点集合 result ，然后我们使用 BFS 的思路从节点 0 开始不断去访问非 restricted 的节点，并不断保存入 result 中，遍历结束我们返回 result 得长度即可。

时间复杂度为 O(N) ，空间复杂度为 O(N) 。

解答

class Solution(object):
    def reachableNodes(self, n, edges, restricted):
        """
        :type n: int
        :type edges: List[List[int]]
        :type restricted: List[int]
        :rtype: int
        """
        d = collections.defaultdict(list)
        for s,t in edges:
            d[s].append(t)
            d[t].append(s)
        restricted = set(restricted)
        result = set()
        result.add(0)
        stack = [0]
        while stack:
            cur = stack.pop(0)
            for sub in d[cur]:
                if sub not in result and sub not in restricted:
                    result.add(sub)
                    stack.append(sub)
        return len(result)

运行结果

62 / 62 test cases passed.
Status: Accepted
Runtime: 1518 ms
Memory Usage: 74.3 MB

原题链接

leetcode.com/contest/wee…

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