leetcode 每日一题 19. 删除链表的倒数第N个节点

Posted 2021-02-21 nil_f

两次遍历

思路:

先遍历一次得到数组长度length，第二次遍历找到位置在length-n的节点p，让p.next=p.next.next即可

代码：

 

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

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        result = ListNode(0)
        result.next = head
        length = 0
        first = head
        while first:
            first = first.next
            length +=1
        length -= n
        first = head 
        for i in range(length-1):
            first = first.next
        first.next = first.next.next
        return result.next

 

快慢指针

两个指针距离为n，同时移动，只需要一次遍历，当快指针到终点，慢指针对应位置，做删除节点操作。

代码：

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

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        result = ListNode(0)
        result.next,fast,slow = head,result,result
        for i in range(n):
            fast = fast.next
        while fast:
            fast = fast.next
            slow = slow.next if fast else slow
        slow.next = slow.next.next
        return result.next