給一個(gè)圖中的n個(gè)節(jié)點(diǎn), 記為 1 到 n . 在開始的時(shí)候圖中沒有邊。
你需要完成下面兩個(gè)方法:
- connect(a, b), 添加連接節(jié)點(diǎn) a, b 的邊.
- query(a, b), 檢驗(yàn)兩個(gè)節(jié)點(diǎn)是否聯(lián)通
樣例
5 // n = 5
query(1, 2) 返回 false
connect(1, 2)
query(1, 3) 返回 false
connect(2, 4)
query(1, 4) 返回 true
代碼
public class ConnectingGraph {
private int[] father = null;
private int find(int x) {
if (father[x] == x) {
return x;
}
return father[x] = find(father[x]);
}
// 初始化時(shí)父結(jié)點(diǎn)都是自己
public ConnectingGraph(int n) {
// initialize your data structure here.
father = new int[n + 1];
for (int i = 1; i <= n; ++i)
father[i] = i;
}
// 連接兩結(jié)點(diǎn)即為并查集合并操作
public void connect(int a, int b) {
int root_a = find(a);
int root_b = find(b);
if (root_a != root_b)
father[root_a] = root_b;
}
// 查詢兩結(jié)點(diǎn)是否相連,即為并查集查詢兩結(jié)點(diǎn)是否擁有相同父結(jié)點(diǎn)
public boolean query(int a, int b) {
int root_a = find(a);
int root_b = find(b);
return root_a == root_b;
}
}