leetcode547

public class Solution {
    private void dfs(int[,] M, int[] visited, int i)
        {
            for (int j = 0; j < M.GetLength(1); j++)
            {
                if (M[i, j] == 1 && visited[j] == 0)
                {
                    visited[j] = 1;
                    dfs(M, visited, j);
                }
            }
        }
        public int FindCircleNum(int[,] M)
        {
            int[] visited = new int[M.GetLength(0)];
            int count = 0;
            for (int i = 0; i < M.GetLength(0); i++)
            {
                if (visited[i] == 0)
                {
                    dfs(M, visited, i);
                    count++;
                }
            }
            return count;
        }
}

这是第一种方法，主要是使用dfs来实现。

下面是第二种方法，是同学用python写的，我翻译成了C#，这种思路是union find。

public class Solution {
    public int FindCircleNum(int[,] M)
        {
            var N = M.GetLength(0);
            var groups = N;
            var leads = new int[N];
            for (var i = 0; i < N; i++)
            {
                leads[i] = i;
            }

            for (var i = 0; i < N; i++)
            {
                for (var j = i + 1; j < N; j++)
                {
                    if (M[i, j] == 1)
                    {
                        var lead1 = find(i, leads);
                        var lead2 = find(j, leads);
                        if (lead1 != lead2)
                        {
                            leads[lead1] = lead2;
                            groups--;
                        }
                    }
                }
            }
            return groups;
        }

        private int find(int x, int[] parents)
        {
            if (parents[x] == x)
            {
                return x;
            }
            else
            {
                return find(parents[x], parents);
            }
        }
}

从耗时角度来看，第一种效率略高一些。

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