【python面试题】链表反转

单链表反转【python实现】

单链表反转方法

头节点开始遍历

1. 用三个指针来解决链表反转，cur是现在的链表节点，tmp暂存cur.next，new_head指向之前的cur
2. 其中，tmp是保存之前的链条不能断的，很重要

尾节点开始递归

1. 递归反转链表，递归：从尾节点指向上一级节点，上一级节点指向None，重复这个过程，直到递归结束

代码如下:

class ListNode:

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


class Solution:
    
    def reverse_list(self, head: ListNode) -> ListNode:
        if head == None or head.next == None:
            return head

        # 用三个指针来解决链表反转，cur是现在的链表节点，tmp暂存cur.next，new_head指向之前的cur 
        cur = head
        tmp = None
        new_head = None
        while cur:
            tmp = cur.next
            cur.next = new_head
            new_head = cur
            cur = tmp

        return new_head       
    
    def reverse_list2(self, head: ListNode) -> ListNode:
        if head == None or head.next == None:
            return head
        # 通过Node指向之前的head来进行链表反转
        Node = None
        while head:
            p = head
            head = head.next
            p.next = Node
            Node = p
        return Node
    
    def reverse_list_recusive(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        # 递归反转链表，递归：从下一级的节点指向上一级节点，上一级节点指向None，重复这个过程，直到递归结束
        NewHead = self.reverse_list_recusive(head.next)
        head.next.next = head
        head.next = None
        return NewHead
        

if __name__ == "__main__":
    n1 = ListNode(1)
    n2 = ListNode(2)
    n3 = ListNode(3)
    n4 = ListNode(4)

    n1.next = n2
    n2.next = n3
    n3.next = n4

    new_head = Solution().reverse_list(n1)
    print(new_head.val)
    print(new_head.next.val)
    print(new_head.next.next.val)
    print(new_head.next.next.next.val)

    print()

    new_head = Solution().reverse_list2(new_head)
    print(new_head.val)
    print(new_head.next.val)
    print(new_head.next.next.val)
    print(new_head.next.next.next.val)

    print()

    new_head = Solution().reverse_list_recusive(new_head)
    print(new_head.val)
    print(new_head.next.val)
    print(new_head.next.next.val)
    print(new_head.next.next.next.val)

THE END

​ 这道题很经常的出现在各个厂的笔试题和面试题中，平时大家估计很少用链表，但是面试经常考，所以也要注意下链表的性质，和怎么反转链表