203.remove-linked-list-elements

Statement

简体中文English

Metadata

• Link: 移除链表元素
• Difficulty: Easy
• Tag: 递归 链表

给你一个链表的头节点 head 和一个整数 val ，请你删除链表中所有满足 Node.val == val 的节点，并返回 新的头节点 。

 

示例 1：

输入：head = [1,2,6,3,4,5,6], val = 6
输出：[1,2,3,4,5]

示例 2：

输入：head = [], val = 1
输出：[]

示例 3：

输入：head = [7,7,7,7], val = 7
输出：[]

 

提示：

• 列表中的节点数目在范围 [0, 104] 内
• 1 <= Node.val <= 50
• 0 <= val <= 50

Metadata

• Link: Remove Linked List Elements
• Difficulty: Easy
• Tag: Recursion Linked List

Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

 

Example 1:

Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]

Example 2:

Input: head = [], val = 1
Output: []

Example 3:

Input: head = [7,7,7,7], val = 7
Output: []

 

Constraints:

• The number of nodes in the list is in the range [0, 104].
• 1 <= Node.val <= 50
• 0 <= val <= 50

Solution

Python 3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeElements(self, head: ListNode, val: int) -> ListNode:
        rt = ListNode()
        res = rt

        while head:
            if head.val != val:
                rt.next = ListNode()
                rt = rt.next
                rt.val = head.val
            head = head.next

        return res.next

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最后更新: October 11, 2023