Leetcode - Reverse Linked List

Question:
Reverse a singly linked list.

My code

/**  iteration
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null)
            return null;
        doubleLinkedList doubleHead = new doubleLinkedList(head.val);
        ListNode temp = head;
        while (temp.next != null) {
            doubleHead.next = new doubleLinkedList(temp.next.val);
            doubleHead.next.prev = doubleHead;
            doubleHead = doubleHead.next;
            temp = temp.next;
        }
        
        ListNode reverseListHead = new ListNode(doubleHead.val);
        ListNode reverseTemp = reverseListHead;
        while (doubleHead.prev != null) {
            reverseTemp.next = new ListNode(doubleHead.prev.val);
            doubleHead = doubleHead.prev;
            reverseTemp = reverseTemp.next;
        }
        
        return reverseListHead;
    }
    
    private class doubleLinkedList {
        int val;
        doubleLinkedList next;
        doubleLinkedList prev;
        doubleLinkedList(int x) { val = x;}
    }
}

/** reversion
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null)
            return null;
        ListNode tail = head;
        while (tail.next != null)
            tail = tail.next;
        reverseListNode(head);
        return tail;
    }
    
    private void reverseListNode(ListNode node) {
        if (node.next == null)
            return;
        reverseListNode(node.next);
        node.next.next = node;
        node.next = null;
        
    }
    
    public static void main(String[] argv) {
        ListNode node1 = new ListNode(1);
        ListNode node2 = new ListNode(2);
        ListNode node3 = new ListNode(3);
        ListNode node4 = new ListNode(4);
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        node4.next = null;
        Solution test = new Solution();
        ListNode head = test.reverseList(node1);
        while (head != null) {
            System.out.println(head.val);
            head = head.next;
        }
    }
    
}
  

My test result:
iteration:

Paste_Image.png

reversion:

Paste_Image.png

**
总结：
这次作业比较简单。用两种方式实现，一种是循环，一种是递归。
循环的思想比较简单，我是先另建一个双向链表，然后把单向链表存到双向链表，之后再把双向链表存到一个新的单向链表中，并返回。或者也可以用数组来实现，把链表结点存入数组，并且不断地resize，存入所有结点，在倒序构造一个新的单向链表。
递归的话，就是把这个链表当做两个部分，当前结点和当前结点之后的那段链表。先把之后的那段链表倒置，再和这个结点倒置。然后不断得递归。
作业还是比较简单的，还是要坚持每天一题。
毕业季，好冷清啊。希望今晚可以闹起来。
**

差不多的做法。
Anyway, Good luck, Richardo！ -- 08/15/2016