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:
Input:n = 5
andedges = [[0, 1], [1, 2], [3, 4]]
0 3 | | 1 --- 2 4 Output: 2
Example 2:
Input:n = 5
andedges = [[0, 1], [1, 2], [2, 3], [3, 4]]
0 4 | | 1 --- 2 --- 3 Output: 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
.
struct Solution;
struct UnionFind {
parent: Vec<usize>,
n: usize,
}
impl UnionFind {
fn new(n: usize) -> Self {
let parent = (0..n).collect();
UnionFind { parent, n }
}
fn find(&mut self, i: usize) -> usize {
let j = self.parent[i];
if i == j {
i
} else {
let k = self.find(j);
self.parent[i] = k;
k
}
}
fn union(&mut self, i: usize, j: usize) {
let i = self.find(i);
let j = self.find(j);
if i != j {
self.parent[i] = j;
self.n -= 1;
}
}
}
impl Solution {
fn count_components(n: i32, edges: Vec<Vec<i32>>) -> i32 {
let n = n as usize;
let mut uf = UnionFind::new(n);
for edge in edges {
let i = edge[0] as usize;
let j = edge[1] as usize;
uf.union(i, j);
}
uf.n as i32
}
}
#[test]
fn test() {
let n = 5;
let edges = vec_vec_i32![[0, 1], [1, 2], [3, 4]];
let res = 2;
assert_eq!(Solution::count_components(n, edges), res);
let n = 5;
let edges = vec_vec_i32![[0, 1], [1, 2], [2, 3], [3, 4]];
let res = 1;
assert_eq!(Solution::count_components(n, edges), res);
}