[LeetCode] 323. Number of Connected Components in an Undirected Graph #323

grandyang · 2019-05-30T16:19:26Z

Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph.

Example 1:

0 3

| |

1 --- 2 4

Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], return 2.

Example 2:

0 4

| |

1 --- 2 --- 3

Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]], return 1.

Note:

You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

这道题让我们求无向图中连通区域的个数，LeetCode中关于图Graph的题屈指可数，解法都有类似的特点，都是要先构建邻接链表Adjacency List来做。这道题的一种解法是利用DFS来做，思路是给每个节点都有个flag标记其是否被访问过，对于一个未访问过的节点，我们将结果自增1，因为这肯定是一个新的连通区域，然后我们通过邻接链表来遍历与其相邻的节点，并将他们都标记成已访问过，遍历完所有的连通节点后我们继续寻找下一个未访问过的节点，以此类推直至所有的节点都被访问过了，那么此时我们也就求出来了连通区域的个数。

解法一：

class Solution {
public:
    int countComponents(int n, vector<pair<int, int> >& edges) {
        int res = 0;
        vector<vector<int> > g(n);
        vector<bool> v(n, false);
        for (auto a : edges) {
            g[a.first].push_back(a.second);
            g[a.second].push_back(a.first);
        }
        for (int i = 0; i < n; ++i) {
            if (!v[i]) {
                ++res;
                dfs(g, v, i);
            }
        }
        return res;
    }
    void dfs(vector<vector<int> > &g, vector<bool> &v, int i) {
        if (v[i]) return;
        v[i] = true;
        for (int j = 0; j < g[i].size(); ++j) {
            dfs(g, v, g[i][j]);
        }
    }
};

这道题还有一种比较巧妙的方法，不用建立邻接链表，也不用DFS，思路是建立一个root数组，下标和节点值相同，此时root[i]表示节点i属于group i，我们初始化了n个部分 (res = n)，假设开始的时候每个节点都属于一个单独的区间，然后我们开始遍历所有的edge，对于一条边的两个点，他们起始时在root中的值不相同，这时候我们我们将结果减1，表示少了一个区间，然后更新其中一个节点的root值，使两个节点的root值相同，那么这样我们就能把连通区间的所有节点的root值都标记成相同的值，不同连通区间的root值不相同，这样也能找出连通区间的个数。

解法二：

class Solution {
public:
    int countComponents(int n, vector<pair<int, int> >& edges) {
        int res = n;
        vector<int> root(n);
        for (int i = 0; i < n; ++i) root[i] = i;
        for (auto a : edges) {
            int x = find(root, a.first), y = find(root, a.second);
            if (x != y) {
                --res;
                root[y] = x;
            }
        }
        return res;
    }
    int find(vector<int> &root, int i) {
        while (root[i] != i) i = root[i];
        return i;
    }
};

类似题目：

参考资料：

https://leetcode.com/discuss/77308/accepted-dfs-in-c

https://leetcode.com/discuss/77027/c-solution-using-union-find

https://leetcode.com/discuss/76519/similar-to-number-of-islands-ii-with-a-findroot-function

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