删除链表的倒数第 N 个结点python3(leetcode19)

#19. 删除链表的倒数第 N 个结点

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

#快慢指针

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        dummy_head = ListNode(0)
        dummy_head.next = head

        slow, fast = dummy_head, dummy_head
        # 定义快慢指针，快指针先走n后，快慢指针一起走，
        # 当快指针走到最后一个节点时，也就是fast.next=0时，慢指针此时达到待删除元素的前面
        while(n != 0):
            fast = fast.next
            n -= 1
        while(fast.next != None):
            fast = fast.next
            slow = slow.next
        temp = slow.next
        slow.next = temp.next

        return dummy_head.next # return的是dummy_head的next

#先求链表长度

# 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: ListNode, n: int) -> ListNode:
        dummy_head = ListNode(0)
        dummy_head.next = head
        curr = dummy_head
        L = 0
        while(head != None): # 使用head遍历，虽然会改变head，但是之前已经将head赋给dummy_head
            head = head.next
            L += 1

        for _ in range(L - n):
            curr = curr.next
        curr.next = curr.next.next
        
        return dummy_head.next