LeetCode 547. Friend Circles 朋友圈

题目

https://leetcode.com/problems/friend-circles/description/

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:

  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.

讲解

用来判断元素在不在同一个集合、合并集合的并查集 Union Find 对于这套题来说是再合适不过了。

当满足 (M[i][j] == 1 && i != j)条件时候,我们通过并查集的union()把两个朋友圈合并。

视频教学

Java参考代码

class UnionFind {
    private int[] parents;
    private int circleCount; 
    
    public UnionFind(int n) {
        parents = new int[n];
        for (int i = 0; i < n; i++) {
            parents[i] = i;
        }
    }
    
    public int find(int x) {
        if (parents[x] == x) {
            return x;
        }
        
        return parents[x] = find(parents[x]);
    }
    
    public void union(int a, int b) {
        int groupA = find(a);
        int groupB = find(b);
        
        if (groupA != groupB) {
            parents[groupA] = groupB;
            circleCount--;
        }        
    }
    
    public void setCircleCount(int circleCount) {
        this.circleCount = circleCount;
    }
    
    public int getCircleCount() {
        return this.circleCount;    
    }    
}

class Solution {
    public int findCircleNum(int[][] M) {
        if (M.length == 0 || M[0].length == 0) {
            return 0;
        }
        
        int m = M.length;
        int n = M[0].length;
        
        UnionFind unionFind = new UnionFind(m * n);
        unionFind.setCircleCount(m);
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (M[i][j] == 1 && i != j) {
                    unionFind.union(i, j);        
                }                
            }            
        }
        
        return unionFind.getCircleCount();
    }
}

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