剑指offer-python 24 链表反转

递归

class Solution:
    # 返回ListNode
    def ReverseList(self, pHead):
        # write code here
        if pHead == None or pHead.next==None:
            return pHead
        pre = pHead.next
        newhead = self.ReverseList(pre)
        pHead.next = None
        pre.next = pHead
        
        return newhead
           

迭代

使用头插法。

class Solution:
    # 返回ListNode
    def ReverseList(self, pHead):
        # write code here
        if pHead == None or pHead.next==None:
            return pHead
        newhead = None
        curhead = pHead
        while curhead:
            tmp = curhead.next
            curhead.next = newhead
            newhead = curhead
            curhead = tmp
        return newhead