题解 | #删除链表的节点#
删除链表的节点
http://www.nowcoder.com/practice/f9f78ca89ad643c99701a7142bd59f5d
#coding:utf-8
class ListNode:
def init(self, x):
self.val = x
self.next = None
代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
@param head ListNode类
@param val int整型
@return ListNode类
class Solution: def deleteNode(self , head , val ): # write code here cur = head pre = ListNode(None) pre.next = cur preHead = pre while cur: if cur.val == val: pre.next = cur.next break else: pre = cur cur = cur.next
return preHead.next
算法太卷啦
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