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:
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Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], return 2.
Example 2:
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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.
思路：并查集(Union find)
class Solution {
public:
int getFather(vector<int>& father, int i) {
if (father[i] == i) return i;
father[i] = getFather(father, father[i]);
return father[i];
}
void merge(vector<int>& father, int i, int j) {
int fatherI = getFather(father, i);
int fatherJ = getFather(father, j);
father[fatherJ] = fatherI;
}
int countComponents(int n, vector<pair<int, int>>& edges) {
vector<int> father;
for (int i = ; i < n; i++) father.push_back(i);
for (int i = , n = edges.size(); i < n; i++)
merge(father, get<>(edges[i]), get<>(edges[i]));
unordered_set<int> unions;
for (int i = ; i < n; i++) unions.insert(getFather(father, i));
return unions.size();
}
};