python_ACM模式《剑指offer刷题》链表1

题目：

面试tips：

询问面试官是否可以改变链表结构

思路：

翻转链表，再遍历链表打印。
想要实现先遍历后输出，即先进后出，因此可借助栈结构。
可用隐式的栈结构，递归来实现。

代码实现：

python 复制代码

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

class myClass:
    def printList(self, head):
        # 前后指针翻转链表
        prev, curr = None, head
        while curr:
            tmp = curr.next
            curr.next = prev
            prev, curr = curr, tmp
        # 翻转后prev即指向新链表的头节点
        # 打印链表
        curr = prev
        while curr:
            print(curr.val)
            curr = curr.next

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 None
    curr = head
    for i in range(1, len(arr)):
        curr.next = ListNode(arr[i])
        curr = curr.next
    # 执行函数
    a = myClass()
    a.printList(head)

python 复制代码

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

class myClass:
    def printList(self, head):
        # 定义一个栈，用来存储遍历过的链表节点
        stack = []
        curr = head
        while 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 None
    curr = head
    for i in range(1, len(arr)):
        curr.next = ListNode(arr[i])
        curr = curr.next
    # 执行函数
    a = myClass()
    a.printList(head)

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

python 复制代码

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

class 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 None
    curr = head
    for i in range(1, len(arr)):
        curr.next = ListNode(arr[i])
        curr = curr.next
    # 执行函数
    a = myClass()
    a.printList(head)