203. 移除链表元素：用Python优雅地解决难题

2023-12-13 08:42:59

def remove_elements(head: ListNode, val: int) -> ListNode:
    """
    Remove all elements from a linked list of integers that have value val.
    
    Args:
        head (ListNode): The head of the linked list.
        val (int): The value to remove.
    
    Returns:
        ListNode: The head of the linked list after removing all elements with value val.
    """
    
    # Initialize a dummy node to simplify the code.
    dummy = ListNode(0)
    dummy.next = head
    
    # Initialize the current node to the dummy node.
    current = dummy
    
    # Iterate over the linked list.
    while current.next is not None:
        
        # If the next node has the value we want to remove, skip it.
        if current.next.val == val:
            current.next = current.next.next
        
        # Otherwise, move the current node forward.
        else:
            current = current.next
    
    # Return the head of the linked list after removing all elements with value val.
    return dummy.next

# Example usage
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
head.next.next.next.next = ListNode(5)

val = 2

# Remove all elements with the value 2 from the linked list.
new_head = remove_elements(head, val)

# Print the linked list after removing all elements with the value 2.
current = new_head
while current is not None:
    print(current.val)
    current = current.next

简介

链表是一种常用的数据结构，广泛应用于各种编程任务中。当需要存储和处理有序数据时，链表通常是一个不错的选择。然而，在某些情况下，我们可能需要从链表中删除具有特定值的元素。这就是LeetCode 203. 移除链表元素问题所要解决的难题。

算法

为了解决这个问题，我们可以使用以下算法：

初始化一个哨兵节点（dummy node）并将其指向链表的头节点。哨兵节点简化了代码，使我们可以将链表视为一个循环链表。
使用一个指针current指向哨兵节点。
循环遍历链表，直到current.next为None。
如果current.next.val等于要删除的值，则将current.next指向current.next.next，跳过具有该值的节点。
否则，将current移到下一个节点。
返回哨兵节点的下一个节点作为新的头节点。

代码实现

def remove_elements(head, val):
    """
    Remove all elements from a linked list of integers that have value val.
    
    Args:
        head (ListNode): The head of the linked list.
        val (int): The value to remove.
    
    Returns:
        ListNode: The head of the linked list after removing all elements with value val.
    """
    dummy = ListNode(0)
    dummy.next = head
    current = dummy
    
    while current.next is not None:
        if current.next.val == val:
            current.next = current.next.next
        else:
            current = current.next
    
    return dummy.next