代码随想录算法训练营day4|● 24. 两两交换链表中的节点 ● 19.删除链表的倒数第N个节点 ● 142.环形链表I

学习资料：https://programmercarl.com/0024.两两交换链表中的节点.html

学习记录：
24.两两交换链表中的节点（添加虚拟头节点；交换1、2节点和3、4节点时，要用1前面的cur，先保存1为temp且3保存为temp1，cur指向2，再把2指向temp，因为cur指向2后就与1没关联了）

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy_head = ListNode(next=head)
        cur=dummy_head
        while cur.next and cur.next.next:
            temp=cur.next
            temp1=cur.next.next.next
            cur.next=cur.next.next
            cur.next.next=temp
            temp.next=temp1
            cur=cur.next.next
        return dummy_head.next

19.删除链表的倒数第N个节点（用快慢指针，fast和slow都设置为虚拟头节点，fast先走N+1步，之后slow和fast同时走，当fast走到NULL，此时slow为倒数第N-1个节点；把slow的下一个节点删除即可）

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy_head=ListNode(next=head)
        slow=fast=dummy_head
        for i in range(n+1):
            fast=fast.next
        while fast:
            fast=fast.next
            slow=slow.next
        slow.next=slow.next.next
        return dummy_head.next

142.环形链表（链表中，若快（2）慢（1）指针能相遇，则为环形链表；从相遇地点到环形入口的距离和从头节点到环形入口的时间一样；相遇时，快指针走了x+y+n(y+z),慢指针走了x+y，则2(x+y)=x+y+n(y+z)，其实不太理解？？）

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

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        fast=slow=head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
            if fast == slow:
                index1 = fast
                index2 = head
                while index1 != index2:
                    index1 = index1.next
                    index2 = index2.next
                return index1
        return None