[LeetCode.547]Friend Circles

问题描述：
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.

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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.

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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.

解题思路：
使用并查集，将朋友关系的学生归为一个集合，用一维数组存储朋友关系，比如1和2是朋友关系，则将2指向1，3和4是朋友关系，则将4指向3。如再有2和4是朋友关系，找到2的祖先（即1），再找到4的祖先（即3），将3指向1。祖先的指针是-1。最后结果如数组所示

1	2	3	4
-1	1	1	3

最后，祖先的个数（即-1的个数）即是最后集合的个数

代码：

public class Solution {
    public int findCircleNum(int[][] M) {
        if(M == null || M.length == 0 || M[0].length == 0){
            return 0;
        }
        int n = M.length;
        int[] set = new int[n];//定义并查集
        Arrays.fill(set,-1);//初始化
        for(int i = 1; i < n; i++){
            for(int j = 0; j < i; j++){//遍历下三角
                if(M[i][j] == 1){
                    join(set,i,j);//找到i和j的祖先，并将一个祖先的上级指向另一个祖先
                }
            }
        }
        int result = 0;
        for(int num : set){
            if(num == -1){//祖先上级值是-1，输出祖先的个数
                result++;
            }
        }
        return result;
    }
    public int find(int[] set,int i){//查找i的祖先
        int result = i;
        while(set[result] != -1){
            result = set[result];
        }
        int j = i;
        int temp;
        while(set[j] != -1){//路径压缩，将之前搜索过的路径的节点的上级都指向新的祖先
            temp = set[j];
            set[j] = result;
            j = temp;
        }
        return result;
    }
    public void join(int[] set,int i,int j){//找到i和j的祖先，并将一个祖先的上级指向另一个祖先
        int x = find(set,i);
        int y = find(set,j);
        if(x != y){
            set[x] = y;
        }
    }
}