[LeetCode328] Odd Even Linked List

最新推荐文章于 2024-08-08 06:51:18 发布

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本文介绍了一种在单链表中将所有奇数位置的节点分组到一起，然后跟随其后放置所有偶数位置节点的方法。此操作保持了奇数和偶数节点内部的相对顺序，并且是在原地进行的，空间复杂度为O(1)，时间复杂度为O(n)。

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:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->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 h1 = head, h2 = head.next;
        ListNode p1 = head, p2 = head.next;
        while(p2 != null && p2.next!= null) {
            p1.next = p2.next;
            p2.next = p2.next.next;
            p1 = p1.next;
            p2 = p2.next;
        }
        p1.next = h2;
        return h1;
    }