LeetCode | 0547. Friend Circles朋友圈【Python】

LeetCode 0547. Friend Circles朋友圈【Medium】【Python】【DFS】

Problem

LeetCode

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:

1. N is in range [1,200].
2. M[i][i] = 1 for all students.
3. If M[i][j] = 1, then M[j][i] = 1.

問題

力扣

班上有 N 名學生。其中有些人是朋友，有些則不是。他們的友誼具有是傳遞性。如果已知 A 是 B 的朋友，B 是 C 的朋友，那麼我們可以認爲 A 也是 C 的朋友。所謂的朋友圈，是指所有朋友的集合。

給定一個 N * N 的矩陣 M，表示班級中學生之間的朋友關係。如果M[i][j] = 1，表示已知第 i 個和 j 個學生互爲朋友關係，否則爲不知道。你必須輸出所有學生中的已知的朋友圈總數。

示例 1:

輸入: 
[[1,1,0],
 [1,1,0],
 [0,0,1]]
輸出: 2 
說明：已知學生0和學生1互爲朋友，他們在一個朋友圈。
第2個學生自己在一個朋友圈。所以返回2。

示例 2:

輸入: 
[[1,1,0],
 [1,1,1],
 [0,1,1]]
輸出: 1
說明：已知學生0和學生1互爲朋友，學生1和學生2互爲朋友，所以學生0和學生2也是朋友，所以他們三個在一個朋友圈，返回1。

注意：

1. N 在[1,200]的範圍內。
2. 對於所有學生，有M[i][i] = 1。
3. 如果有M[i][j] = 1，則有M[j][i] = 1。

思路

DFS

求無向圖連通塊的個數。
訪問過打上標記。

時間複雜度: O(n^2)，需要遍歷 M 矩陣。
空間複雜度: O(n)，visited 數組的大小。

Python3代碼

class Solution:
    def findCircleNum(self, M: List[List[int]]) -> int:
        m = len(M)
        ans, visited = 0, set()

        # def template
        def dfs(i):
            for j in range(m):
                if M[i][j] and j not in visited:  # 1 and not visited
                    visited.add(j)
                    dfs(j)

        for i in range(m):
            if i not in visited:  # not visited
                dfs(i)
                ans += 1
        return ans

代碼地址

GitHub鏈接