LinkedList - 142. Linked List Cycle II
142. Linked List Cycle II
Given a linked list, return the node where the cycle begins. If there is no cycle, return null
.
To represent a cycle in the given linked list, we use an integer pos
which represents the position (0-indexed) in the linked list where tail connects to. If pos
is -1
, then there is no cycle in the linked list.
Note: Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: tail connects to node index 1 Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0 Output: tail connects to node index 0 Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1 Output: no cycle Explanation: There is no cycle in the linked list.
Follow-up: Can you solve it without using extra space?
思路:
题目是让找到环的起点,做法就是用快慢指针,开始的时候,两个指针都从head出发,然后快指针走两步,慢指针一次走一步,相遇就说明有环,这个时候,再用慢指针从链表头开始走,快指针变成走一步,这样两个指针再次相遇的地方就是环的入口。
代码:
java:
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
if (head == null || head.next == null) return null;
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (fast == slow) {
while (head != fast) {
fast = fast.next;
head = head.next;
}
return head;
}
}
return null;
}
}
go:
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func detectCycle(head *ListNode) *ListNode {
var (
res *ListNode
slow = head
fast = head
)
if head == nil || head.Next == nil {
return res
}
for fast != nil && fast.Next != nil {
slow = slow.Next
fast = fast.Next.Next
if slow == fast {
fast = head
for slow != fast {
slow = slow.Next
fast = fast.Next
}
return slow
}
}
return res
}
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