從尾到頭打印鏈表

輸入一個鏈表的頭節點，從尾到頭反過來返回每個節點的值 （用數組返回）。示例 1：輸入：head = [1,3,2] 輸出：[2,3,1]

提供三種題解

# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = Noneclass Solution:"""解題思路一: 遞歸解法1.f(-r) = f(head.next) + head.val2.f(head.next) 為遞歸函數解題思路二: 利用棧的特性，先入后出1.遍歷鏈表，不停的入棧2.不停的彈出棧，并把值放到數組中解題思路三: 1.先翻轉整個鏈表2.再順序輸出"""# 解題思路一def reversePrint(self, head: ListNode) -> List[int]:result = list()self.reverseTrave(head, result)return resultdef reverseTrave(self, head, result):if not head:return self.reverseTrave(head.next, result)result.append(head.val)# 解題思路二def reversePrint(self, head: ListNode) -> List[int]:stack = list()result = list()# 遍歷入棧while head:stack.append(head.val)head = head.next# 遍歷出棧while stack:result.append(stack.pop())return result# 解題思路三def reversePrint(self, head: ListNode) -> List[int]:new_head = Noneresult = list()# 翻轉鏈表while head:per = head.nexthead.next = new_headnew_head = headhead = per# 遍歷翻轉后的鏈表while new_head:result.append(new_head.val)new_head = new_head.nextreturn result

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