LeetCode 237. Delete Node in a Linked List(删除链表中的结点)
题目描述
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
Supposed the linked list is 1 -> 2 -> 3 -> 4
and you are given the third node with value 3
, the linked list should become 1 -> 2 -> 4
after calling your function.
解法一
思路
注意这题没有限定结点的值不能修改,而且直接给出需要删除的结点本身,而不是其值。那么这就有一个猥琐的办法了:
- 将当前结点值修改为【后一结点】值;
- 删除【后一结点】。
以1 -> 2 -> 3 -> 4
中删除3
为例:
1 -> 2 -> 4 -> 4
;1 -> 2 -> 4
。
回头想想,为什么要绕一圈删除【后一结点】而不是当前结点本身呢?因为题目中给的链表是单链表,当前结点只有next
指针指向其【后一结点】,即只能修改当前结点之后的链表结构,所以为了简便起见,就拿【后一结点】开刀了。
Python
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def deleteNode(self, node):
"""
:type node: ListNode
:rtype: void Do not return anything, modify node in-place instead.
"""
node.val = node.next.val
node.next = node.next.next
Java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public void deleteNode(ListNode node) {
node.val = node.next.val;
node.next = node.next.next;
}
}
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void deleteNode(ListNode *node) {
node->val = node->next->val;
node->next = node->next->next;
}
};