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.

1.用dfs

class Solution {
public:
  int countComponents(int n, vector<pair<int, int> >& edges) {
    vector<vector<int> > nl(n);
    for (auto p : edges){
      nl[p.first].push_back(p.second);
      nl[p.second].push_back(p.first);
    }
    vector<int> visited(n, false);
    int cnt = 0;
    for (int i = 0; i < n; i++){
      if (!visited[i]){
        cnt++;
        dfs(nl, visited, i);
      }
    }
    return cnt;
  }
private:
  void dfs(vector<vector<int> >& nl, vector<int>&visited, int i){
    if (visited[i]) return;
    visited[i] = true;
    for (int v : nl[i]){
      if (!visited[v]){
        dfs(nl, visited, v);
      }
    }
  }
};

2.用并查集

class Solution {
public:
  int countComponents(int n, vector<pair<int, int> >& edges) {
    vector<int> s(n);
    for (int i = 0; i < n; i++){
      s[i] = i;
    }
    for (auto e : edges){
      int ra = find(s, e.first);
      int rb = find(s, e.second);
      if (ra != rb){
        s[ra] = rb;
      }
    }
    int cnt = 0;
    for (int i = 0; i < n; i++){
      if (s[i] == i){
        cnt++;
      }
    }
    return cnt;
  }
private:
  int find(vector<int>&s, int x){
    int r = x;
    while (r != s[r]){
      r = s[r];
    }

    int j = x;
    while (s[j] != j){
      int tmp = s[j];
      s[j] = r;
      j = tmp;
    }
    
    return r;
  }
};