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.
这道题和numbers of islands II 是一个思路，一个count初始化为n，union find每次有新的edge就union两个节点，如果两个节点(u, v)原来不在一个连通图里面就减少count并且连起来，如果原来就在一个图里面就不管。用一个索引array来做，union find优化就是加权了，每次把大的树的root当做parent，小的树的root作为child。
Java:
public class Solution {
public int countComponents(int n, int[][] edges) {
int count = n;
// array to store parent
init(n, edges);
for(int[] edge : edges) {
int root1 = find(edge[0]);
int root2 = find(edge[1]);
if(root1 != root2) {
union(root1, root2);
count--;
}
}
return count;
}
int[] map;
private void init(int n, int[][] edges) {
map = new int[n];
for(int[] edge : edges) {
map[edge[0]] = edge[0];
map[edge[1]] = edge[1];
}
}
private int find(int child) {
while(map[child] != child) child = map[child];
return child;
}
private void union(int child, int parent) {
map[child] = parent;
}
}
