leetcode142 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.
 """
 """
 判断链表是否有环。有两种做法
 第一种是用set()存已经遍历过的结点
 如果新的结点在set()里，则返回，有环
 """

 class ListNode:
     def __init__(self, x):
         self.val = x
         self.next = None

 class Solution:
     def detectCycle(self, head):
         nodes = set()
         while head:
             if head in nodes:
                 return head
             nodes.add(head)
             head = head.next
         return None

 """
 解法二：快慢指针
 A→B→C       快指针：A, C, B, D 一次走两步
    ↖↓       慢指针：A, B, C, D
     D
 根据距离推算：相遇点距环入口的距离 = (头节点距环入口的距离)*（快指针步数-1)
 快慢指针相遇在D，让快指针变为head
 同步向后，再次相遇即为环的起点
 """

 class ListNode:
     def __init__(self, x):
         self.val = x
         self.next = None

 class Solution:
     def detectCycle(self, head):
         if head == None or head.next == None:
             return None
         slow, fast = head, head
         while fast:
             slow = slow.next
             fast = fast.next
             if fast:
                 fast = fast.next
                 if fast == slow:
                     break
         if slow != fast:
             return None
         fast = head #!!!关键所在
         while slow:
             if fast == slow:
                 return fast
             fast = fast.next
             slow = slow.next