OCI 面试经验分享：链表节点反转 | 面试代面, OA代做, 面试辅导

OCI 面试经验分享：链表节点反转 | 面试代面, OA代做, 面试辅导

在这次OCI的面试中，面试官给出了一道关于链表节点反转的题目，要求考生实现一个算法来反转链表中每两个相邻的节点。这道题目考察了考生对链表的基本操作以及对指针操作的熟练程度。在有限的时间内写出正确且高效的代码，对考生的编码能力和算法理解是一个很大的考验。

如果你想在这样的面试中取得好成绩，我们的代面服务能够帮助你深入理解算法，提升编码能力，并在面试中自信地展示你的技能。

想要了解更多或获取我们的服务，欢迎添加微信 leetcode-king。

---

题目：反转链表中的每两个相邻节点

Problem Description:

Write a function to reverse every two adjacent nodes in a linked list. You should not modify the values in the list's nodes, only the nodes themselves may be changed.

Example:

Given a linked list: 1->2->3->4->5, the function should return 2->1->4->3->5.

Code:

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
        next = null;
    }
}

public class Solution {
    public ListNode reverseEvery2Nodes(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        ListNode cur = head;

        while (cur != null && cur.next != null) {
            ListNode nextPair = cur.next.next;
            ListNode second = cur.next;

            // Reverse the pair
            second.next = cur;
            cur.next = nextPair;
            prev.next = second;

            // Update pointers
            prev = cur;
            cur = nextPair;
        }

        return dummy.next;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(4);
        head.next.next.next.next = new ListNode(5);

        ListNode result = solution.reverseEvery2Nodes(head);
        while (result != null) {
            System.out.print(result.val + " ");
            result = result.next;
        }
    }
}