LeetCode328 奇偶链表

最新推荐文章于 2025-03-15 00:15:00 发布

原创最新推荐文章于 2025-03-15 00:15:00 发布 · 525 阅读

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本文介绍了一种在单链表中将所有奇数节点聚集在一起，随后跟上所有偶数节点的方法。此操作需在原地进行，且程序应在O(1)的空间复杂度和O(nodes)的时间复杂度下运行。文章提供了具体的代码实现，展示了如何通过遍历链表并调整指针来完成这一任务。

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Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL
Note:

The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on …

public ListNode oddEvenList(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode evenStart = head.next;
        ListNode even = evenStart;
        ListNode odd = head;
        while(even != null && even.next != null){
            odd.next = odd.next.next;
            even.next = even.next.next;
            odd = odd.next;
            even = even.next;
        }
        odd.next = evenStart;
        return head;
    }