剑指 Offer 18. 删除链表的节点

给定单向链表的头指针和一个要删除的节点的值，定义一个函数删除该节点。

返回删除后的链表的头节点。

示例一：

输入: head = [4,5,1,9], val = 5
输出: [4,1,9]
解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 4 -> 1 -> 9.

示例二：

输入: head = [4,5,1,9], val = 1
输出: [4,5,9]
解释: 给定你链表中值为 1 的第三个节点，那么在调用了你的函数之后，该链表应变为 4 -> 5 -> 9.

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

class Solution():
    def deleteNode(self, head: ListNode, val: int) -> ListNode:
        if head.val == val: return head.next
        temp = head
        while temp.next is not None:
            if temp.next.val is not None:
                temp.next = temp.next.next
                break
            temp = temp.next
        return head

node3 = ListNode()
node3.val = 3
node3.next = None

node2 = ListNode()
node2.val = 2
node2.next = node3

node1 = ListNode()
node1.val = 1
node1.next = node2

s = Solution()
head = s.deleteNode(node1, 2)
print(head)