Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
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題意：查詢環形連結串列的環的起點
思路：類似於141，快慢指標法可判斷環的存在與否
設頭結點到環的起點距離為a，環的起點到第一次相遇點距離為b，第一次相遇點到環的起點距離為c，則第一次相遇快指標走了（a+b+c+b）,慢指標走了（a+b）,快指標速度是慢指標的兩倍，則可得a=c。所以將快指標置於連結串列開頭，慢指標仍在相遇點，二者同時同速走，則相遇點就是環的起點
注意判斷兩指標是否相等，不是判斷兩指標的值是否相等
Runtime: 76 ms

# 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):
        dummy=ListNode(0)
        dummy.next=head
        slow=fast=dummy
        while fast.next and
 
 fast.next.next:
            slow=slow.next
            fast=fast.next.next
            if slow==fast:
                fast=dummy
                while slow!=fast:
                    slow=slow.next
                    fast=fast.next
                return slow 
        return  None