LeetCode 547. Friend Circles--Python解法--笔试算法题

题目地址：Friend Circles - 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:
N is in range [1,200].
M[i][i] = 1 for all students.
If M[i][j] = 1, then M[j][i] = 1.

这道题目我有印象是一家公司的笔试算法题，意思是无向图找分成几个部分，几个环。

当时的Python解法如下：

class Solution:def findCircleNum(self, M: List[List[int]]) -> int:num = len(M)s = {i: set() for i in range(0, num)}for i in range(0, num):for j in range(0, num):if M[i][j] == 1:s[i].add(j)stack = set([0])passed = set()result = 1while stack != set():temp=stack.pop()passed.add(temp)for i in s[temp]:if i not in passed:stack.add(i)if stack==set():if len(passed)==num:breakelse:result+=1stack.add(list(set(s)-passed)[0])return result

上面的解法肯定不是最优的，但已经可以过所有样例了。
具体的思路是从一个人开始，进行BFS遍历，遍历完后找没遍历过的人开始BFS，一直到结束。
然后我看了别人的解答，发现并没有比我的做法好很多。。。。

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LeetCode 547. Friend Circles--Python解法--笔试算法题

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