subject ：
Given the head pointer of one-way linked list and the value of a node to be deleted , Define a function to delete the node .
Return the head node of the deleted linked list .
Be careful ： This question is different from the original one
Example 1:
Input : head = [4,5,1,9], val = 5
Output : [4,1,9]
explain : Given that the value of your list is 5 Second node of , So after calling your function , The list should be 4 -> 1 -> 9.
Example 2:
Input : head = [4,5,1,9], val = 1
Output : [4,5,9]
explain : Given that the value of your list is 1 The third node of , So after calling your function , The list should be 4 -> 5 -> 9.
code:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def deleteNode(self, head: ListNode, val: int) -> ListNode:
temp = head
if temp.val == val:
return temp.next
while temp.next:
if temp.next.val == val:
temp.next = temp.next.next
else:
temp = temp.next
return head
