题目：

面试tips：

询问面试官是否可以改变链表结构

思路：

1. 翻转链表，再遍历链表打印。

2. 想要实现先遍历后输出，即先进后出，因此可借助栈结构。

3. 可用隐式的栈结构，递归来实现。

代码实现：

1.

class ListNode:def __init__(self, val = 0, next = None):self.val = valself.next = nextclass myClass:def printList(self, head):# 前后指针翻转链表prev, curr = None, headwhile curr:tmp = curr.nextcurr.next = prevprev, curr = curr, tmp# 翻转后prev即指向新链表的头节点# 打印链表curr = prevwhile curr:print(curr.val)curr = curr.nextif __name__ == '__main__':# 构造测试用例 -- 用数组构造链表arr = [5,4,2,3]head = ListNode(arr[0]) if arr else Nonecurr = headfor i in range(1, len(arr)):curr.next = ListNode(arr[i])curr = curr.next# 执行函数a = myClass()a.printList(head)

2.

class ListNode:def __init__(self, val = 0, next = None):self.val = valself.next = nextclass myClass:def printList(self, head):# 定义一个栈，用来存储遍历过的链表节点stack = []curr = headwhile curr:stack.append(curr)curr = curr.next# 打印链表值，这里pop出来也可释放内存while stack:node = stack.pop()print(node.val)if __name__ == '__main__':# 构造测试用例 -- 用数组构造链表arr = [5,4,2,3]head = ListNode(arr[0]) if arr else Nonecurr = headfor i in range(1, len(arr)):curr.next = ListNode(arr[i])curr = curr.next# 执行函数a = myClass()a.printList(head)

3.

采用递归的思想 注意是递归到最后一个元素才开始打印 即要先写递归 后写打印代码

class ListNode:def __init__(self, val = 0, next = None):self.val = valself.next = nextclass myClass:def printList(self, head):# 递归打印链表 -- 递归就是栈 也就相当于使用了一个隐式的栈结构# 终止条件if not head:return# 单层递归逻辑 -- 先递归后打印self.printList(head.next)print(head.val)if __name__ == '__main__':# 构造测试用例 -- 用数组构造链表arr = [5,2,6,3,5,4]head = ListNode(arr[0]) if arr else Nonecurr = headfor i in range(1, len(arr)):curr.next = ListNode(arr[i])curr = curr.next# 执行函数a = myClass()a.printList(head)