题解 | #删除链表的倒数第n个节点#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param n int整型 
# @return ListNode类
#
# 左右双指针，右指针领先左指针 n-1 个位置
# 需要判断一个特殊情况：链表长度等于 n 时，此时需要删除头节点
# 节点向右移动时需要保存左节点的前一个节点
class Solution:
    def removeNthFromEnd(self , head: ListNode, n: int) -> ListNode:
        l, r = head, head
        
        for i in range(n-1):
            r = r.next
        if r.next == None:
            return head.next
        
        while r.next != None:
            temp = l
            l = l.next
            r = r.next
        temp.next = l.next

        return head