题目信息

• 链接: remove-linked-list-elements

• 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, 10<sup>4</sup>].
  • 1 <= Node.val <= 50
  • 0 <= val <= 50

• 描述:

  • 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。

解题思路

• 遍历链表，找到值等于val的节点v的前一个节点，删除链表元素v 删除节点
• 因为可能会删除头节点所以使用 dummy node
• 需要删除所有值等于val的节点，所以需要使用while循环持续删除节点
• 注意获取 p.val 时p是否可能为None

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 removeElements(self, head: ListNode, val: int) -> ListNode:
		  dummy = ListNode()
		  dummy.next = head
		  p = dummy
		  while p != None and p.next != None:
			  while p.next != None and p.next.val == val:
				  p.next = p.next.next
			  p = p.next
		  return dummy.next