547. Friend Circles

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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:

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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:

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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1： DFS

这里因为已经有一个矩阵代表了graph连接的情况，所以我们不需要另外建图了。
对于每一个node，dfs一下把所有连通的地方都设为visited。
在外面循环的时候，每次看到一个没有被visited过得就是一个新的group

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class Solution {
    public int findCircleNum(int[][] M) {
        if (M == null || M.length == 0 || M[0] == null || M[0].length == 0) {
            return 0;
        }
        int n = M.length;
        boolean[] visited = new boolean[n];
        int res = 0;
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                res++;
                dfs(M, visited, i);
            }
        }
        return res;
    }
    private void dfs(int[][] M, boolean[] visited, int id) {
        visited[id] = true;
        for (int i = 0; i < M.length; i++) {
            if (visited[i] || M[i][id] == 0) continue;
            dfs(M, visited, i);
        }
    }
}

解法2: Union Find

比较典型的Union Find解法。

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class Solution {
    public int findCircleNum(int[][] M) {
        if (M == null || M.length == 0 || M[0] == null || M[0].length == 0) {
            return 0;
        }
        int n = M.length;
        int[] roots = new int[n];
        for (int i = 0; i < n; i++) {
            roots[i] = i;
        }
        // loop through M to union
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1 ; j < n; j++) {
                if (M[i][j] == 1) {
                    int root_left = find(roots, i);
                    int root_right = find(roots, j);
                    roots[root_right] = root_left;
                }
            }
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
            if (roots[i] == i) {
                res++;
            }
        }
        return res;
    }
    private int find(int[] nums, int i) {
        while (i != nums[i]) {
            i = nums[nums[i]];
        }
        return i;
    }
}