删除链表中给定位置的节点

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def deleteNode(head, position):
    if position < 0:
        return head

    if position == 0:
        return head.next

    curr = head
    for _ in range(position - 1):
        curr = curr.next
        if curr is None:
            return head

    if curr.next is None:
        return head

    curr.next = curr.next.next
    return head

# 示例用法
# 创建链表: 1 -> 2 -> 3 -> 4 -> 5
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)

# 删除位置为2的节点
new_head = deleteNode(head, 2)

# 输出链表: 1 -> 2 -> 4 -> 5
curr = new_head
while curr:
    print(curr.val)
    curr = curr.next