[LeetCode] Reverse Linked List 倒置链表,Reverse Linked List II

 

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

 

之前做到 Reverse Linked List II 的时候我还纳闷怎么只有二没有一呢,原来真是忘了啊,现在才加上,这道题跟之前那道比起来简单不少,题目为了增加些许难度,让我们分别用迭代和递归来实现,但难度还是不大。我们先来看迭代的解法,思路是在原链表之前建立一个空的newHead,因为首节点会变,然后从head开始,将之后的一个节点移到newHead之后,重复此操作直到head成为末节点为止,代码如下:

 

解法一:

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *newHead = NULL;
        while (head) {
            ListNode *t = head->next;
            head->next = newHead;
            newHead = head;
            head = t;
        }
        return newHead;
    }
};

 

下面我们来看递归解法,代码量更少,递归解法的思路是,不断的进入递归函数,直到head指向倒数第二个节点,因为head指向空或者是最后一个结点都直接返回了,newHead则指向对head的下一个结点调用递归函数返回的头结点,此时newHead指向最后一个结点,然后head的下一个结点的next指向head本身,这个相当于把head结点移动到末尾的操作,因为是回溯的操作,所以head的下一个结点总是在上一轮被移动到末尾了,但head之后的next还没有断开,所以可以顺势将head移动到末尾,再把next断开,最后返回newHead即可,代码如下:

 

解法二:

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode *newHead = reverseList(head->next);
        head->next->next = head;
        head->next = NULL;
        return newHead;
    }
};

 

类似题目:

Reverse Linked List II

Binary Tree Upside Down

Palindrome Linked List

 

参考资料:

https://leetcode.com/problems/reverse-linked-list/

https://leetcode.com/problems/reverse-linked-list/discuss/58156/My-Java-recursive-solution

https://leetcode.com/problems/reverse-linked-list/discuss/58337/Fast-Recursive-Java-solution

https://leetcode.com/problems/reverse-linked-list/discuss/58125/In-place-iterative-and-recursive-Java-solution

 

    原文作者:Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/4478820.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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