思路:遍歷鏈表,將節點存入Set,遇到重複節點即返回。
import java.util.*;
public class Solution {
public ListNode detectCycle(ListNode head) {
if(head == null){
return null;
}
Set set = new HashSet();
set.add(head);
while(head.next!=null){
if(set.contains(head.next)){
return head.next;
}else{
set.add(head.next);
head=head.next;
}
}
return null;
}
}
補充:不用額外空間的解法
/**
* 題目描述: 鏈表的入環節點,如果無環,返回null
* Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.
* Follow up: Can you solve it without using extra space?
* 思路:
* 1)首先判斷是否有環,有環時,返回相遇的節點,無環,返回null
* 2)有環的情況下, 求鏈表的入環節點
* fast再次從頭出發,每次走一步,
* slow從相遇點出發,每次走一步,
* 再次相遇即爲環入口點。
* 注:此方法在牛客BAT算法課鏈表的部分有講解。
*/
//nowcoder pass
public class Solution {
public ListNode detectCycle(ListNode head) {
if (head == null) {
return null;
}
ListNode meetNode = meetingNode(head);
if (meetNode == null) {//說明無環
return null;
}
ListNode fast = head;
ListNode slow = meetNode;
while (slow != fast) {
slow = slow.next;
fast = fast.next;
}
return slow;
}
//尋找相遇節點,如果無環,返回null
public ListNode meetingNode(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
return slow;
}
}
return null;
}
}