剑指offer13题-在O(1)时间内删除链表节点

1、输入是val: ListNode 解法

# 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: ListNode) -> ListNode:
        if not (head and val):
            return False
        if val.next != None:
            tmp = val.next
            val.value = tmp.value
            val.next = tmp.next
        elif head == val:
            head = None
        else:
            node = head
            while node.next!=val:
                node = node.next
            node.next = None
        return head

2、输入是 val: int 解法

class Solution:
    def deleteNode(self, head: ListNode, val: int) -> ListNode:
        dummy = ListNode(0)  # 设置伪结点
        dummy.next = head
        if head.val == val: return head.next # 头结点是要删除的点，直接返回
        while head and head.next:
            if head.next.val == val:   # 找到了要删除的结点，删除
                head.next = head.next.next
            head = head.next   # 找到了之后为什么还要移动一个
        return dummy.next