0203.Remove_Linked_List_Elements.py

"""
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
"""
# 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: Optional[ListNode], val: int) -> Optional[ListNode]:
        if not head:
            return None
        
        dummyNode = ListNode(-1, head)
        prev, curr = dummyNode, head
        while curr:
            if curr.val == val:    #这里不能加上prev = prev.next 因为后面一个curr也有可能是val
                prev.next = curr.next
                curr = curr.next
            else:
                prev = prev.next
                curr = curr.next
                
        return dummyNode.next