142. Linked List Cycle II

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LeetCode

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Problem

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?

Code

如何判断有没有环？如何得到进入环的节点？

class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;

        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast){
                break;
            }
        }

        if(fast == null || fast.next == null) return null;

        fast = head;
        while(slow != fast){
            fast = fast.next;
            slow = slow.next;
        }

        return slow;
    }
}

写法2，与题目287类似

class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode slow = dummy;
        ListNode fast = dummy;

        while(true){
            if(fast == null || fast.next == null) break;
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast){
                break;
            }
        }

        if(fast == null || fast.next == null) return null;

        fast = dummy;
        while(true){
            fast = fast.next;
            slow = slow.next;
            if(slow == fast){
                break;
            }
        }

        return slow;
    }
}