更新时间:2022-02-08 04:18:39
请编写一个函数,使其可以删除某个链表中给定的(非末尾)节点。传入函数的唯一参数为 要被删除的节点 。
现有一个链表 -- head = [4,5,1,9],它可以表示为:
输入:
head = [4,5,1,9], node = 5
输出:
[4,1,9]
解释:
给定你链表中值为 5 的第二个节点,那么在调用了你的函数之后,该链表应变为 4 -> 1 -> 9.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public void deleteNode(ListNode node) {
node.val = node.next.val;
node.next = node.next.next;
}
}
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
void deleteNode(struct ListNode* node) {
node->val = node->next->val;
node->next = node->next->next;
}
/**
* 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;
}
};
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
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
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func deleteNode(node *ListNode) {
node.Val = node.Next.Val;
node.Next = node.Next.Next;
}
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