【LeetCode Practice】[Medium]547. Number of Provinces
547. Number of Provinces
Algorithm idea: array
topic:
It is equivalent to finding how many connected components in this graph, you can use DFS, BFS, and check the set.
Record DFS BFS as follows
java code-DFS
class Solution {
public int findCircleNum(int[][] isConnected) {
int num = isConnected.length;
int count = 0;
boolean[] visit = new boolean[num];//用来标记节点是否访问过
for (int i = 0; i < num; i++) {
if (!visit[i]) {
visit[i] = true;//标记当前节点访问过
myDfs(isConnected, visit, num, i);//DFS向下查找一次,则连通分量加一
count++;//连通分量加一
}
}
return count;
}
private void myDfs(int[][] isConnected, boolean[] visit, int num, int node) {
for (int i = 0; i < num; i++) {
//查找相邻节点,是否能到达
if (!visit[i] && isConnected[node][i] == 1) {
//如果相邻能连通
visit[i] = true;//标记当前节点访问过
myDfs(isConnected, visit,num, i);//用相邻节点,向下递归
}
}
}
}
java code-BFS
class Solution {
public int findCircleNum(int[][] isConnected) {
int num = isConnected.length;
int count = 0;
boolean[] visit = new boolean[num];
Queue<Integer> queue = new LinkedList<>();
//开始遍历
for (int i =0 ; i < num; i++){
if(!visit[i]){
queue.offer(i);
visit[i] = true;
while (!queue.isEmpty()) {
int node = queue.poll();
for (int j =0 ; j < num; j++){
if(!visit[j] && isConnected[node][j] == 1){
queue.offer(j);
visit[j] = true;
}
}
}
count++;
}
}
return count;
}
}
java code-and check set
class Solution {
public int findCircleNum(int[][] isConnected) {
int provinces = isConnected.length;
int[] parent = new int[provinces];
for (int i = 0; i < provinces; i++) {
parent[i] = i;
}
for (int i = 0; i < provinces; i++) {
for (int j = i + 1; j < provinces; j++) {
if (isConnected[i][j] == 1) {
union(parent, i, j);
}
}
}
int circles = 0;
for (int i = 0; i < provinces; i++) {
if (parent[i] == i) {
circles++;
}
}
return circles;
}
public void union(int[] parent, int index1, int index2) {
parent[find(parent, index1)] = find(parent, index2);
}
public int find(int[] parent, int index) {
if (parent[index] != index) {
parent[index] = find(parent, parent[index]);
}
return parent[index];
}
}