判断图是否连通三种方法 dfs bfs 并查集（代码带更新）

数据结构 图 判断图是否连通
图的遍历方法有dfs，bfs先遍历一遍图，判断是否所有的节点都遍历到

并查集的方法合并节点，最后数一共有多少集合，如果集合是1，证明连通。
参考这篇博客http://www.cnblogs.com/noKing/p/8018609.html

LeetCode547 Friend Circles朋友圈 代码 好理解版本

【思路】灵感来源于上面的参考博客，先生产一维数组p，数组的值和下标一致。遍历二维数组，给每个值找一个老大。相当于形成了一条条链表。
如0和1是朋友，将0->1
0和2是朋友 发现p[0]已经不是0了，说明0有老大了，去找他的老大，发现是1，因为p[1]==1，将p[1]置为2。现在0->1->2了。
安装上面思路编写代码，发现如下问题。
以下代码再运行[[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1]]用例时出错，死循环。
原因在于执行一下代码出现问题。假设第一行0和1，2，3，4已经合并完。按照我的代码的意思
0->1->2->3->4,遍历到第2行的时候，1要和2合并，这时候1会一直找到4，发现4==p[4]，把p[4]置为2，这里逻辑就出错了。正确的应该判断1和2是不是已经有一个相同的根节点了。

while (node != p[node]) {

                        node = p[node];
                    }
                    p[node] = j;

问题代码

public class Solution {
    public static void main(String[] args) {
        int[][] M = {{1,1,1,1,1},{1,1,1,1,1},{1,1,1,1,1},{1,1,1,1,1},{1,1,1,1,1}};
        System.out.println(findCircleNum(M));
    }
    public static int findCircleNum(int[][] M) {
        int m = M.length;
        int[] p = new int[m];
        for (int i = 0; i < m; i++) {
            p[i] = i;
        }
        for (int i = 0; i < m; i++) {
            for (int j = i+1; j < m; j++) {
                if (M[i][j] == 1) {
                    int node = p[i];
                    while (node != p[node]) {

                        node = p[node];
                    }
                    p[node] = j;
                }
            }
        }
        //遍历p，看还有多少个i=p[i]，满足i=p[i]的节点都是顶点
        int group = 0;
        for (int i = 0; i < m; i++) {
            if (i == p[i]) {
                ++group;
            }
        }
        return group;
    }
}

改进方法，每次联合两个数的时候先判断是否有共同的根节点。
跑如下测试用例出错

[[1,1,0,0,0,0,0,0,0,0],[1,1,0,0,0,1,0,1,0,0],[0,0,1,0,1,0,0,1,0,0],[0,0,0,1,0,0,1,0,0,0],[0,0,1,0,1,0,1,0,0,0],[0,1,0,0,0,1,0,1,1,0],[0,0,0,1,1,0,1,0,0,0],[0,1,1,0,0,1,0,1,0,0],[0,0,0,0,0,1,0,0,1,0],[0,0,0,0,0,0,0,0,0,1]]
输出：
3
预期：
2

int[][] M ={{1,1,0,0,0,0,0,0,0,0},
                    {1,1,0,0,0,1,0,1,0,0},
                    {0,0,1,0,1,0,0,1,0,0},
                    {0,0,0,1,0,0,1,0,0,0},
                    {0,0,1,0,1,0,1,0,0,0},
                    {0,1,0,0,0,1,0,1,1,0},
                    {0,0,0,1,1,0,1,0,0,0},
                    {0,1,1,0,0,1,0,1,0,0},
                    {0,0,0,0,0,1,0,0,1,0},
                    {0,0,0,0,0,0,0,0,0,1}};

debug发现是break错了，给为continue就过了

package 牛客刷题.LeetCode.并查集.朋友圈;

/**
 * Created by Administrator on 2018/5/28 0028.
 */
public class Solution {
    public static void main(String[] args) {
        //int[][] M = {{1,1,1,1,1},{1,1,1,1,1},{1,1,1,1,1},{1,1,1,1,1},{1,1,1,1,1}};
        int[][] M ={{1,1,0,0,0,0,0,0,0,0},
                    {1,1,0,0,0,1,0,1,0,0},
                    {0,0,1,0,1,0,0,1,0,0},
                    {0,0,0,1,0,0,1,0,0,0},
                    {0,0,1,0,1,0,1,0,0,0},
                    {0,1,0,0,0,1,0,1,1,0},
                    {0,0,0,1,1,0,1,0,0,0},
                    {0,1,1,0,0,1,0,1,0,0},
                    {0,0,0,0,0,1,0,0,1,0},
                    {0,0,0,0,0,0,0,0,0,1}};
        System.out.println(findCircleNum(M));
    }
    public static int findCircleNum(int[][] M) {
        int m = M.length;
        int[] p = new int[m];
        for (int i = 0; i < m; i++) {
            p[i] = i;
        }
        for (int i = 0; i < m; i++) {
            for (int j = i+1; j < m; j++) {
                if (M[i][j] == 1) {
                    int firstRoot = find(i, p);
                    int secondRoot = find(j, p);
                    if (firstRoot == secondRoot) {
                        break;
                    } else {
                        p[firstRoot] = j;
                    }
                }
            }
        }
        //遍历p，看还有多少个i=p[i]，满足i=p[i]的节点都是顶点
        int group = 0;
        for (int i = 0; i < m; i++) {
            if (i == p[i]) {
                ++group;
            }
        }
        return group;
    }

    //找每个节点的根节点下标
    public static int find(int a,int[] p) {
        int node = a;
        while (node != p[node]) {
            node = p[node];
        }
        return node;
    }

}

通过代码

package 牛客刷题.LeetCode.并查集.朋友圈;

/**
 * Created by Administrator on 2018/5/28 0028.
 */
public class Solution {
    public static void main(String[] args) {
        //int[][] M = {{1,1,1,1,1},{1,1,1,1,1},{1,1,1,1,1},{1,1,1,1,1},{1,1,1,1,1}};
        int[][] M ={{1,1,0,0,0,0,0,0,0,0},
                    {1,1,0,0,0,1,0,1,0,0},
                    {0,0,1,0,1,0,0,1,0,0},
                    {0,0,0,1,0,0,1,0,0,0},
                    {0,0,1,0,1,0,1,0,0,0},
                    {0,1,0,0,0,1,0,1,1,0},
                    {0,0,0,1,1,0,1,0,0,0},
                    {0,1,1,0,0,1,0,1,0,0},
                    {0,0,0,0,0,1,0,0,1,0},
                    {0,0,0,0,0,0,0,0,0,1}};
        System.out.println(findCircleNum(M));
    }
    public static int findCircleNum(int[][] M) {
        int m = M.length;
        int[] p = new int[m];
        for (int i = 0; i < m; i++) {
            p[i] = i;
        }
        for (int i = 0; i < m; i++) {
            for (int j = i+1; j < m; j++) {
                if (M[i][j] == 1) {
                    int firstRoot = find(i, p);
                    int secondRoot = find(j, p);
                    if (firstRoot == secondRoot) {
                        continue;
                    } else {
                        p[firstRoot] = j;
                    }
                }
            }
        }
        //遍历p，看还有多少个i=p[i]，满足i=p[i]的节点都是顶点
        int group = 0;
        for (int i = 0; i < m; i++) {
            if (i == p[i]) {
                ++group;
            }
        }
        return group;
    }

    //找每个节点的根节点下标
    public static int find(int a,int[] p) {
        int node = a;
        while (node != p[node]) {
            node = p[node];
        }
        return node;
    }
}