Title description ( portal )
There are n cities, some of which are connected to each other, and some are not connected. If city a and city b are directly connected, and city b is directly connected to city c, then city a and city c are indirectly connected.
A province is a group of directly or indirectly connected cities. The group does not contain other unconnected cities.
Give you an nxn matrix isConnected, where isConnected[i][j] = 1 means that the ith city and the jth city are directly connected, and isConnected[i][j] = 0 means that the two are not directly connected.
Returns the number of provinces in the matrix.
Example 1:
输入：isConnected = [[1,1,0],[1,1,0],[0,0,1]]
输出：2
Example 2:
输入：isConnected = [[1,0,0],[0,1,0],[0,0,1]]
输出：3
Problem solving ideas & code implementation
This question is similar to: [One question every day] Niu Ke: Calculating Moments (Consolidated Checking Exercise 01)
Knowledge Points: [High-level Data Structure] Consolidating Detailed Explanations
/**
* @ClassName lk547
* @Description ：TODO
* @Author Josvin
* @Date 2021/01/16/12:01
*/
public class lk547 {
private static int count;
private static int[] id;
private static int[] size;
public int findCircleNum(int[][] isConnected) {
Union_Find(isConnected);
return count;
}
private void Union_Find(int[][] isConnected) {
int N = isConnected.length;
id = new int[N];
size = new int[N];
count = N;
for (int i = 0; i < N; i++) {
id[i] = i;
size[i] = 1;
}
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (1 == isConnected[i][j]) {
Union(i, j);
}
}
}
}
private static void Union(int p, int q) {
int proot = Find(p);
int qroot = Find(q);
if (proot == qroot) {
return;
}
if (size[p] > size[q]) {
id[proot] = qroot;
size[qroot] += size[p];
} else {
id[qroot] = proot;
size[proot] += size[q];
}
count--;
}
private static int Find(int p) {
if (id[p] != p) {
id[p] = Find(id[p]);
}
return id[p];
}
}