• [LeetCode] Number of Connected Components in an Undirected Graph 无向图中的连通区域的个数


    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:

    Clone Graph

    Minimum Height Trees 

    Course Schedule

    Course Schedule II

    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

    LeetCode All in One 题目讲解汇总(持续更新中...)

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  • 原文地址:https://www.cnblogs.com/grandyang/p/5166356.html
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