142 Linked List Cycle II

返回循环开始的节点： 使用快慢指针。

Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        slow, fast = head, head
        while True:
            if fast is None:
                return None
            if fast.next and fast.next.next:
                slow = slow.next
                fast = fast.next.next
            else:
                return None
            if fast == slow:
                fast = head
                while fast != slow:
                    fast = fast.next
                    slow = slow.next
                return fast

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) {
        ListNode slow = head, fast = head;
        while(true){
            if(fast == null){
                return null;
            }
            if (fast.next != null && fast.next.next != null){
                slow = slow.next;
                fast = fast.next.next;
            }
            else{
                return null;
            }
            if (fast == slow){
                fast = head;
                while(fast != slow){
                    fast = fast.next;
                    slow = slow.next;
                }
                return fast;
            }
        }
    }
}