朋友圈 題目 解析

class Solution {

    class UF {
        private int count;
        private int parent[]; //它的父節點
        private int size[]; //以自己為根節點的樹有多少節點

        public UF(int n) {
            this.count = n;
            parent = new int[n];
            size = new int[n];
            //初始化的時候自己的父節點指向自己
            for
 
 (int i = 0; i < n; i++) {
                parent[i] = i;
                size[i] = 1; 
            }
        }

        //找到x的根節點
        private int find(int x) {
            while (x != parent[x]) {
                //路徑壓縮，畫個圖就明白了。就是和自己的父親做兄弟。
                parent[x] = parent[parent[x]];
                x 
 
= parent[x];
            }
            return x;
        }


        //連線
        public void union(int p, int q) {
            //找到各自的根節點。
            int rootP = find(p); //你看，每次find的時候都進行了路徑壓縮。
            int rootQ = find(q);
            if (rootP == rootQ) return;

            //把小的樹接到大的樹上面
            if
 
 (size[rootQ] > size[rootP]) {
                parent[rootP] = rootQ;
                size[rootQ] += size[rootP];
            } else {
                parent[rootQ] = rootP;
                size[rootP] += size[rootQ];
            }
            count--;
        }

        public boolean connect(int p, int q) {
            return find(p) == find(q);
        }

        public int count() {
            return count;
        }
    }

    public int findCircleNum(int[][] M) {
        int N = M.length;
        UF uf = new UF(N);
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                if (M[i][j] == 1) {
                    uf.union(i, j);
                }
            }
        }
        return uf.count();
    }
}

深度優先

class Solution {
    public int findCircleNum(int[][] M) {
        int N = M.length, count = 0;
        boolean[] V = new boolean[N];
        for (int i = 0; i < N; i++) {
            if (!V[i]) {
                count++;
                dfs(V, M, i);
            }
        }
        return count;
    }

    private void dfs(boolean[] V, int[][]M, int i) {
        for (int j = 0; j < M.length; j++) {
            if (M[i][j] == 1 && !V[j]) {
                V[j] = true;
                dfs(V, M, j);
            }
        }
    }
}

廣度優先

class Solution {
    public int findCircleNum(int[][] M) {
        int N = M.length, count = 0;
        Queue<Integer> queue = new LinkedList<>();
        boolean[] V = new boolean[N];
        for (int i = 0; i < N; i++) {
            if (!V[i]) {
                queue.offer(i);
                count++;
                while (!queue.isEmpty()){
                    int k = queue.poll();
                    V[k] = true;
                    for (int j = 0; j < N; j++) {
                        if (M[k][j] == 1 && !V[j]) {
                            queue.offer(j);
                        } 
                    }
                }
            }
        }
        return count;
    }
}