leetcode 19. Remove Nth Node From End of List【Python刷题】

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本文介绍了一种算法，用于从链表中删除倒数第N个节点，并保持链表的连续性。通过一次遍历实现，提高了效率。文章提供了Python代码示例，展示了如何使用两个指针来解决此问题。

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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?

注意是如果长度和原来的链表长度一样，要特判，直接return即可

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

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        r=head
        for i in range(n):
            r=r.next
        if(r==None):
            return head.next
        l=head
        while(r.next!=None):
            r=r.next
            l=l.next
        x=l.next
        l.next=x.next
        return head