203. Remove Linked List Elements Given theheadof a linked list and an integerval, remove all the nodes of the linked list that hasNode.val == val, and returnthe new head. Example 1: Input:head = [1,2,6,3,4,5,6], val = 6Output:[1,2,3,4,5] Example 2: Input:head = [],...
Remove all elements from a linked list of integers that have valueval. Example: Input:1->2->6->3->4->5->6,val= 6Output:1->2->3->4->5 说到删除,首先想到定义两个指针,分别指向要被删除的结点和该结点的前驱结点。这里还需要考虑头结点是需要删除结点的特殊情况。 /** * Definition for s...
Remove all elements from a linked list of integers that have valueval. Example Given:1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6,val= 6 Return:1 --> 2 --> 3 --> 4 --> 5 Credits: Special thanks to@mithmattfor adding this problem and creating all test cases. 本题就是...
class Solution { public: ListNode* removeElements(ListNode* head, int val) { if(head==NULL) return NULL; while(head!=NULL&&head->val==val) head=head->next; if(head==NULL) return NULL; ListNode* tem =head; while(tem->next){ if(tem->next->val==val){ tem->next=tem->next->next...
Timeit import Timeit from Utility.playground import inputToListNode, listNodeToString, ListNode """ https://leetcode.cn/problems/remove-linked-list-elements/ """ class Solution1: def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]: dummy = cur = fast = head ...
Remove all elements from a linked list of integers that have value val. Example Given: 1 –> 2 –> 6 –> 3 –> 4 –> 5 –> 6, val = 6 Return: 1 –> 2 –> 3 –> 4 –> 5 解题思路 定义两个指针pre和cur,如果cur的值为val,则删除该结点。需要注意的情况有两种:①需要删除头结...
203. Remove Linked List Elements Remove all elements from a linked list of integers that have valueval. Example Given:1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6,val= 6 Return:1 --> 2 --> 3 --> 4 --> 5 删除链表中指定的所有元素。
Remove all elements from a linked list of integers that have valueval. Example Given:1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6,val= 6 Return:1 --> 2 --> 3 --> 4 --> 5 题目大意: 删除链表中全部的目标元素。 代码如下: ...
func removeElements(head *ListNode, val int) *ListNode { if head == nil { return head } var p1 ListNode p1.Next = head p2 := &p1 for head != nil { if head.Val == val { p2.Next, head = head.Next, head.Next } else { p2, head = head, head.Next } } return p1.Next ...
# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: def removeElements(self, head: ListNode, val: int) -> ListNode: prev = ListNode(0) prev.next = head cur = prev while cur.next...