剑指 Offer 18. 删除链表的节点
本问题对应的 leetcode 原文链接：剑指 Offer 18. 删除链表的节点
问题描述
给定单向链表的头指针和一个要删除的节点的值，定义一个函数删除该节点。
返回删除后的链表的头节点。
注意：此题对比原题有改动
示例 1:
输入: head = [4,5,1,9], val = 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.
示例 2:
输入: head = [4,5,1,9], val = 1
输出: [4,5,9]
解释: 给定你链表中值为 1 的第三个节点，那么在调用了你的函数之后，该链表应变为 4 -> 5 -> 9.
说明：
题目保证链表中节点的值互不相同
若使用 C 或 C++ 语言，你不需要 free 或 delete 被删除的节点
代码实现
public ListNode deleteNode(ListNode head, int val) {
if(head == null){
return null;
}
if(head.val == val){
return head.next;
}
ListNode temp = head.next;
ListNode pre = head;
while(temp != null){
if(temp.val == val){
pre.next = temp.next;
return head;
}
temp = temp.next;
pre = pre.next;
}
return head;
}
时间复杂度：O(n)
空间复杂度：O(1)
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, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
if head == None:
return None
if head.val == val:
return head.next
temp = head.next
pre = head
while temp != None:
if temp.val == val:
pre.next = temp.next
return head
temp = temp.next
pre = pre.next
return head
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteNode(ListNode* head, int val) {
if (head == nullptr) {
return nullptr;
}
if (head->val == val) {
return head->next;
}
ListNode* temp = head->next;
ListNode* pre = head;
while (temp != nullptr) {
if (temp->val == val) {
pre->next = temp->next;
return head;
}
temp = temp->next;
pre = pre->next;
}
return head;
}
};
Go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func deleteNode(head *ListNode, val int) *ListNode {
/*
:type head: ListNode
:type val: int
:rtype: ListNode
*/
if head == nil {
return nil
}
if head.Val == val {
return head.Next
}
temp := head.Next
pre := head
for temp != nil {
if temp.Val == val {
pre.Next = temp.Next
return head
}
temp = temp.Next
pre = pre.Next
}
return head
}
JS
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} val
* @return {ListNode}
*/
const deleteNode = (head, val) => {
if(head === null){
return null;
}
if(head.val === val){
return head.next;
}
let temp = head.next;
let pre = head;
while(temp !== null){
if(temp.val === val){
pre.next = temp.next;
return head;
}
temp = temp.next;
pre = pre.next;
}
return head;
};
