【LeetCode】19. 删除链表的倒数第 N 个结点

题目

思想

找到 倒数第 n+1 个节点，然后删除即可。需要特判一下删除的是否是头结点。

代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        length = 0
        tmp = head
        # 求出链表长度
        while(tmp):
            tmp = tmp.next
            length+=1
        # print(length)

        # 倒数第n个节点=> 正数第 length-n+1, 所以遍历到第 length-n 个节点处即可删除
        cnt = 1
        tmp = head
        res = None # 返回的节点
        if (length -n ) == 0: # 删除头结点            
            return head.next
        while(tmp and cnt < length-n):
            tmp = tmp.next
            cnt+=1
        # print(tmp.val)

        # 此时
        invalid_node = tmp.next
        tmp.next = tmp.next.next
        del invalid_node
        return