[leetcode]206. Reverse Linked List

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?

Solution:

非递归和递归实现。

非递归版本：定义三个指针，pre,cur,next。从前向后修改指针。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null ||head.next == null) {
            return head;
        }
        ListNode cur = head;
        ListNode pre = null;
        ListNode next = null;
        ListNode newhead = null;
        while(cur != null) {
            next = cur.next;
            if(next == null) {    //说明到达最后
                newhead = cur;
                cur.next = pre;  //最后一个也要连上前面
                break;
            }
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return newhead;
    }
}

递归版本：和迭代版本思路类似，只不过这个是从后向前修改。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null ||head.next == null) {
            return head;
        }
        ListNode newhead = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return newhead;
    }
}