题目描述:
给定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。
代码:
# 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
"""
slow = fast = head
for i in range(n):
fast = fast.next
if not fast: # 这个判断语句是为了处理一些边界情况
return head.next
while fast.next:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return head
思路分析:
用双指针:slow
和 fast
。参见: https://www.cnblogs.com/yangyalong/p/9744829.html
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