leecode-206-反转链表

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

Related Topics

链表

\n

👍 1696

👎 0

解法1：迭代

//反转一个单链表。 
//
// 示例: 
//
// 输入: 1->2->3->4->5->NULL
//输出: 5->4->3->2->1->NULL 
//
// 进阶: 
//你可以迭代或递归地反转链表。你能否用两种方法解决这道题？ 
// Related Topics 链表 
// 👍 1696 👎 0


//leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        // 迭代 翻转的关键在于将后节点改成前节点
        ListNode prev = null;
        ListNode curr = head;
        while(curr != null){
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
}
//leetcode submit region end(Prohibit modification and deletion)

解法二：递归

//反转一个单链表。 
//
// 示例: 
//
// 输入: 1->2->3->4->5->NULL
//输出: 5->4->3->2->1->NULL 
//
// 进阶: 
//你可以迭代或递归地反转链表。你能否用两种方法解决这道题？ 
// Related Topics 链表 
// 👍 1696 👎 0


//leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
       //递归
        if (head == null || head.next == null) {
            return head;
        }
        ListNode newHead = reverseList(head.next);
        //将后节点的值改成前节点，关键点还有，head的后节点注意置空
        head.next.next = head;
        head.next = null;
        return newHead;
    }
}
//leetcode submit region end(Prohibit modification and deletion)