[LinkedList]019 Remove Nth Node From End of List

  • 分类:LinkedList

  • 考察知识点:LinkedList

  • 最优解时间复杂度:**O(n) **

19. Remove Nth Node From End of List

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

代码:

解法:

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

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        #初始化
        res=ListNode(0)
        #注意这里fast和slow的作用是两个指针
        fast=res
        slow=res
        res.next=head
        
        #让它先走两步
        for i in range(n+1):
            fast=fast.next
            
        while(fast!=None):
            fast=fast.next
            slow=slow.next
        
        if slow==None:
            return res.next
        
        slow.next=slow.next.next
        
        return res.next

讨论:

1.这道题的思路是两个指针你追我赶!然后自己想是想不到的,我还是看了视频解2333333
2.就是觉得ListNode在Python里很奇怪,然后才发现slow和fast其实是两个指针

[LinkedList]019 Remove Nth Node From End of List_第1张图片
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