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
.
Solution 1:
BFS
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class Solution { | |
public: | |
int countComponents(int n, vector<pair<int, int>>& edges) { | |
vector<vector<int> > links(n); | |
for(int i=0; i<edges.size(); i++){ | |
pair<int, int> p = edges[i]; | |
links[p.first].push_back(p.second); | |
links[p.second].push_back(p.first); | |
} | |
vector<int> visited(n, 1); | |
int res=0; | |
for(int i=0; i<links.size(); i++){ | |
if(links[i].size()==0) | |
continue; | |
res++; | |
queue<int> myQ; | |
myQ.push(i); | |
while(!myQ.empty()){ | |
int front = myQ.front(); | |
visited[front]=0; | |
myQ.pop(); | |
for(int j=0; j<links[front].size(); j++){ | |
int p=links[front][j]; | |
if(links[p].size()!=0) | |
myQ.push(p); | |
for(int k=0; k<links[p].size(); k++){ | |
if(links[p][k]==front){ | |
int tmpp = links[p][links[p].size()-1]; | |
links[p][links[p].size()-1]=links[p][k]; | |
links[p][k]=tmpp; | |
links[p].pop_back(); | |
break; | |
} | |
} | |
} | |
while(!links[front].empty()) | |
links[front].pop_back(); | |
} | |
} | |
for(int i=0; i<n; i++) | |
res=res+visited[i]; | |
return res; | |
} | |
}; |
Union Find
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class Solution { | |
public: | |
int find(vector<int>& roots, int id){ | |
while(roots[id] != id){ | |
roots[id] = roots[roots[id]]; | |
id = roots[id]; | |
} | |
return id; | |
} | |
int countComponents(int n, vector<pair<int, int>>& edges) { | |
vector<int> roots(n); | |
for(int i=0; i<n; i++) | |
roots[i]=i; | |
for(int i=0; i<edges.size(); i++){ | |
int root1 = find(roots, edges[i].first); | |
int root2 = find(roots, edges[i].second); | |
if(root1!=root2){ | |
roots[root1]=root2; | |
n--; | |
} | |
} | |
return n; | |
} | |
}; |
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