【启发】leetcode -- 547. Friend Circles【并查集 + 集合标识更新的时机】

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: 2Explanation: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: 1Explanation: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:

1. N is in range [1,200].
2. M[i][i] = 1 for all students.
3. If M[i][j] = 1, then M[j][i] = 1.

题意概述

给定一个矩阵N*N ,表示N个人之间的朋友关系。朋友关系具有传递性，试找出这N个人中有多少个朋友团体。

分析及解答

解法1：（简明、清晰）

• 未进行路径压缩
• 参考：（https://discuss.leetcode.com/topic/85039/java-solution-union-find）

   public int findCircleNum(int[][] M) {        int m = M.length, cnt = 0;        int[] root = new int[m];         for (int i = 0; i < m; i++) root[i] = i;         for (int i = 0; i < m; i++)             for (int j = i + 1; j < m; j++)                if (M[i][j] == 1) unionFind(root, i, j);        for (int i = 0; i < m; i++)            if (i == root[i]) cnt++;        return cnt;    }        void unionFind (int[] root, int v1, int v2) {        while (root[v1] != v1) v1 = root[v1]; //find v1's root        while (root[v2] != v2) v2 = root[v2]; //find v2's root        if (root[v1] != root[v2]) root[v2] = v1; //unite the 2 subtrees     }

解法2：（我的丑的版本）

• 【注意：发现新的朋友关系之后的影响】当在遍历过程中发现新的朋友关系时，我们不仅仅需要考虑其对于当下正在处理的朋友关系的影响，同时还要考虑对于过去已经建立的朋友关系的影响（可能会连接起以前独立的两个朋友团体）。（易忽略）

class Solution {public static int findCircleNum(int[][] M) {if(M == null || M.length == 0){return 0;}int[] friends =  new int[M.length];for(int i = 0; i < friends.length;i++){friends[i] = -1; // -1 表示尚未分配集合。}for(int i = 0; i < M.length; i++){for(int j = i ; j < M.length;j++){//【关键】初始化 朋友关系。（注意遍历次序：自上而下，自左向右）（一个集合中最小的数作为集合的标志）                //只有发现存在朋友关系时，才进行相应的处理。if(M[i][j] == 1){                    //若之前并未分配集合，首次分配集合。if(friends[j] == -1){friends[j] = i;}else{                        //若j之前已经分配了集合。则需要进一步传递这种朋友关系。int newRoot = find(friends, j);                        //因为i中可能有自己的集合标志。而与 j的朋友关系，将这两个集合关联了起来，形成一个集合。所以，                        //我们将其中一个集合的根标记为另一个集合的根。friends[friends[i]] = newRoot;                        //将i的根标记为新的集合的根。friends[i] = newRoot;}}}}int result = 0;for(int i = 0; i < friends.length; i++){            //每个 i = friends[i]的人，都代表着一个朋友团体。if(i == friends[i]){result ++;}}return result;}public static int find(int[] friends,int i){        //查找 i 所属于的集合的根（集合的标志）。while(i != friends[i]){i = friends[i];}return i;}}