【leetcode】【206】Reverse Linked List

一、问题描述

Reverse a singly linked list.

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Hint:

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

二、问题分析

这道题，最先想到的是迭代的方式。而迭代又有两种方式：①保持从头访问过的元素先对位置不变，只改变指针的方向，头结点需要单独处理；②指针方向不变，改变元素的位置，即每访问一个新的元素，就把新元素插到头上；

递归的方法其实是从链表的最后边依次改变指针的方向。

三、Java AC代码

迭代①

public ListNode reverseList(ListNode head) {        if (head==null || head.next == null) {return head;}        ListNode pre = head;        ListNode cur = head.next;        pre.next = null;        ListNode nxt = null;        while(cur!=null){        nxt = cur.next;        cur.next = pre;        pre = cur;        cur = nxt;        }        return pre;    }

迭代②

public ListNode reverseList(ListNode head) {        if (head==null || head.next == null) {return head;}        ListNode pivot = head;        ListNode frontier = null;        while(pivot.next!=null){            frontier = pivot.next;            pivot.next = frontier.next;            frontier.next = head;            head = frontier;        }        return head;    }

递归

public ListNode reverseList(ListNode head) {        if(head == null ||head.next == null){            return head;        }                ListNode root = reverseList(head.next);                head.next.next = head;        head.next = null;        return root;    }