LeetCode 第 547 题：朋友圈（并查集）

地址：https://leetcode-cn.com/problems/friend-circles/

方法一：并查集

Java 代码：

public class Solution {

    private class UnionFind {
        private int[] parent;
        private int count;

        public UnionFind(int n) {
            count = n;
            parent = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
        }

        public int find(int x) {
            while (x != parent[x]) {
                parent[x] = parent[parent[x]];
                x = parent[x];
            }
            return x;
        }

        public void union(int x, int y) {
            int rootX = find(x);
            int rootY = find(y);
            if (rootX == rootY) {
                return;
            }
            count--;
            parent[rootX] = rootY;
        }
    }

    public int findCircleNum(int[][] M) {
        int len = M.length;
        UnionFind unionFind = new UnionFind(len);

        for (int i = 0; i < len; i++) {
            for (int j = i + 1; j < len; j++) {
                if (M[i][j] == 1) {
                    unionFind.union(i, j);
                }
            }
        }
        return unionFind.count;
    }

    public static void main(String[] args) {
        int[][] M = {{1, 1, 0},
                {1, 1, 0},
                {0, 0, 1}};

        Solution solution = new Solution();
        int res = solution.findCircleNum(M);
        System.out.println(res);
    }
}

方法二：深度优先遍历

C++ 代码：

#include <iostream>
#include <vector>

using namespace std;

class Solution {
private:
    /**
     * 记录是否被访问过
     */
    vector<int> visited;

    /**
     * 一共 N 名学生
     */
    int N;

    void dfs(int u, vector<vector<int>> &M) {
        if (visited[u]) {
            return;
        }

        visited[u] = 1;

        for (int i = 0; i < N; ++i) {
            // 如果是朋友关系，就继续遍历下去，这是深度优先遍历
            if (M[u][i]) {
                dfs(i, M);
            }
        }
    }

public:
    int findCircleNum(vector<vector<int>> &M) {
        N = M.size();
        if (N == 0) {
            return 0;
        }

        // 连通分量
        int res = 0;
        visited.assign(N, 0);

        for (int i = 0; i < N; ++i) {
            if (visited[i]) {
                continue;
            }

            dfs(i, M);
            res++;
        }
        return res;
    }
};