题目

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.

A province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.

Return the total number of provinces.

Example 1:

Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2

Example 2:

Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3

Constraints:

• 1 <= n <= 200
• n == isConnected.length
• n == isConnected[i].length
• isConnected[i][j] is 1 or 0.
• isConnected[i][i] == 1
• isConnected[i][j] == isConnected[j][i]

思路

dfs搜索一座城市关联的所有城市，并标记已搜过。

并查集，待研究。

代码

python版本：

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        def fill(i):
            if isConnected[i][i] == 0:
                return
            isConnected[i][i] = 0
            for j in range(len(isConnected[i])):
                if isConnected[i][j]:
                    fill(j)
        cnt = 0
        for i in range(len(isConnected)):
            if isConnected[i][i]:
                cnt += 1
                fill(i)
        return cnt