[leetcode] 547. Friend Circles @ python

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原题

There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.

Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.

Example 1:
Input:
[[1,1,0],
[1,1,0],
[0,0,1]]
Output: 2
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle.
The 2nd student himself is in a friend circle. So return 2.
Example 2:
Input:
[[1,1,0],
[1,1,1],
[0,1,1]]
Output: 1
Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends,
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
Note:
N is in range [1,200].
M[i][i] = 1 for all students.
If M[i][j] = 1, then M[j][i] = 1.

解法

DFS. 构建seen集合储存已找到的朋友成员, 遍历N个学生, 如果学生i不在seen里, 则在M[i]中寻找学生i 的朋友, 并且深度优先搜索, 寻找学生i朋友的朋友, 将所有i 所在圈子里的成员加到seen里.
Time: O(n^2), n是学生的个数
Space: O(1)

代码

class Solution:
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """
        def dfs(i):
            for nei, val in enumerate(M[i]):
                if val == 1 and nei not in seen:
                    seen.add(nei)
                    dfs(nei)
            
        N = len(M)
        seen = set()
        res = 0
        for i in range(N):
            # if student i has not visited in the circle
            if i not in seen:
                dfs(i)
                res += 1
                
        return res