There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.
A province is a group of directly or indirectly connected cities and no other cities outside of the group.
You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.
Return the total number of provinces.
Example 1:
Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]] Output: 2
Example 2:
Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]] Output: 3
Constraints:
1 <= n <= 200n == isConnected.lengthn == isConnected[i].lengthisConnected[i][j]is1or0.isConnected[i][i] == 1isConnected[i][j] == isConnected[j][i]
struct Solution;
impl Solution {
fn dfs(m: &mut Vec<Vec<i32>>, visited: &mut Vec<bool>, i: usize, n: usize) {
for j in 0..n {
if m[i][j] == 1 && !visited[j] {
visited[j] = true;
Self::dfs(m, visited, j, n);
}
}
}
fn find_circle_num(mut m: Vec<Vec<i32>>) -> i32 {
let mut res = 0;
let n = m.len();
let mut visited: Vec<bool> = vec![false; n];
for i in 0..n {
if !visited[i] {
visited[i] = true;
Self::dfs(&mut m, &mut visited, i, n);
res += 1;
}
}
res
}
}
#[test]
fn test() {
let m: Vec<Vec<i32>> = vec_vec_i32![[1, 1, 0], [1, 1, 0], [0, 0, 1]];
assert_eq!(Solution::find_circle_num(m), 2);
let m: Vec<Vec<i32>> = vec_vec_i32![[1, 1, 0], [1, 1, 1], [0, 1, 1]];
assert_eq!(Solution::find_circle_num(m), 1);
}