Link: https://leetcode.cn/problems/number-of-provinces/

Question

difficulty: mid
adj diff: 3

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]

这道题不难，只要考察 bfs 代码的数量城都。

一次过。

Code

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public int findCircleNum(int[][] isConnected) {
    int n = isConnected.length;
    int count = 0;
    boolean[] visited = new boolean[n];
    Queue<Integer> q = new LinkedList<Integer>();

    while (true) {
        // 1. find a un-visited node from all nodes
        for (int i = 0; i < n; i++) {
            if (visited[i] == false) {
                q.add(i);
                count++;
                visited[i] = true;
                break;
            }
        }

        // 2. start BFS using the node above
        if (q.isEmpty()) {
            break;
        } else {
            // mark all adjacent nodes are visited
            while (!q.isEmpty()) {
                int cur = q.remove();
                for (int j = 0; j < n; j++) {
                    if (j != cur && isConnected[j][cur] == 1 && !visited[j]) {
                        q.add(j);
                        visited[j] = true;
                    }
                }
            }
        }
    }
    return count;
}

实际上，两个 loop 可以合二为一。

如下（别人的代码）：

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public int findCircleNum(int[][] isConnected) {
    int cities = isConnected.length;
    boolean[] visited = new boolean[cities];
    int provinces = 0;
    Queue<Integer> queue = new LinkedList<Integer>();
    for (int i = 0; i < cities; i++) {
        if (!visited[i]) {
            queue.offer(i);
            while (!queue.isEmpty()) {
                int j = queue.poll();
                visited[j] = true;
                for (int k = 0; k < cities; k++) {
                    if (isConnected[j][k] == 1 && !visited[k]) {
                        queue.offer(k);
                    }
                }
            }
            provinces++;
        }
    }
    return provinces;
}