【leetcode】206. Reverse Linked List My Submissions Question
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TAT 递归的好难写啊。。先po iterately的
# Definition for singly-linked list.# class ListNode(object):# def __init__(self, x):# self.val = x# self.next = Noneclass Solution(object): def reverseList(self, head): """ :type head: ListNode :rtype: ListNode """ if (head is None) or (head.next is None): return head p1 = head p2 = head.next p3 = p2.next head.next = None while p3 != None: p2.next = p1 p1 = p2 p2 = p3 p3 = p3.next p2.next = p1 return p2
if 判断一开始没写,卡了好久。 噢 然后while循环的结束条件有点不优雅,还是勉强补了个p2.next = p1 。//这句开始忘写了也跪好几次。。。
递归真的好难写啊啊啊 啊。看了discussion里的一个c++答案。嗯 。再复现一下吧。。。。(此刻都快背出答案来了)
噢对了。 【递归】函数可以重新另外写啊蠢孩子。&& 关于要循环迭递归的输入变量&& 一直return 递归结果的 程序 真是好无力啊。。。
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