力扣--19删除链表中的倒数第n个节点

题目

python代码

正序删除

将倒数第n个转换成正数的处理。然后分为头、尾、中三种情况处理：

def removeNthFromEnd( head, n):
    pointer = head
    length = 0
    while pointer:
        pointer = pointer.next
        length += 1
    if length == 1:
        return None
    pointer = head
    if n >= length:
        head.val = head.next.val
        head.next = head.next.next
    else:
        if n == 1:
            for i in range(length - 2):
                pointer = pointer.next
            pointer.next = None
        else:
            for i in range(length - n - 1):
                pointer = pointer.next
            pointer.next = pointer.next.next
    return head

双指针

用两个相差n位的指针遍历链表，用右指针判断，用左指针进行操作。

def removeNthFromEnd(head, n):
    if head.next == None:
        return None
    left = right = head
    si = 0
    while si < n:
        si += 1
        right = right.next
    if right == None:
        head = head.next
        return head
    while right.next != None:
        left = left.next
        right = right.next
    if n == 1:
        left.next = None
    else:
        left.next = left.next.next
    return head