LeetCode #547 Friend Circles

题目

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.

解题思路

这道题很明显就是求无向图中连通分量的数目问题。基本思路就是分别对每个未访问过的结点进行一次DFS遍历，遍历过程中记录已访问过的结点，最后返回DFS遍历的次数，就是 friend circles 的数目。一般的做法是设置一个 visited 数组来记录已经访问过的结点，这需要 O(n) 的空间复杂度。但是我们注意到这个无向图是用

01

矩阵来表示的，我们可以在遍历的过程中直接把已经访问过的路线置为

0

表示已经访问过，（比如，在DFS过程中访问了从

i

结点到

j

结点的路线

M[i][j]

，此时设置

M[i][j] = M[j][i] = 0

即可表示

i

和

j

结点都已经访问过 ，则以后就不会重复访问了）。利用这个算法可以把空间复杂度降到 O(1)。

Java 代码实现

class Solution {
public int findCircleNum(int[][] M) {
int cnt = 0;
for (int i = 0; i < M.length; ++i) {
// 只有当 M[i][i] != 0 时，结点i才是未访问过的
if (M[i][i] != 0) {
cnt += DFS(i, M);
}
}

return cnt;
}

public int DFS(int node, int[][] M) {
for (int i = 0; i < M.length; ++i) {
if (M[node][i] == 1) {
M[node][i] = M[i][node] = 0;
DFS(i, M);
}
}

// 一次DFS遍历结束，返回1
return 1;
}
}