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.
一刷
题解：union and find
在find的过程中compress可以减小树的深度，加快速度
class Solution {
public int countComponents(int n, int[][] edges) {
int[] roots = new int[n];
for(int i=0; i<n; i++){
roots[i] = i;
}
for(int[] edge : edges){
int rootA = find(edge[0], roots);
int rootB = find(edge[1], roots);
if(rootA != rootB){
roots[rootA] = rootB;
n--;
}
}
return n;
}
private int find(int node, int[] roots){
while(node!=roots[node]){
roots[node] = roots[roots[node]];//compression
node = roots[node];
}
return node;
}
}
