LeetCode 328. Odd Even Linked List

最新推荐文章于 2025-02-03 15:37:10 发布

原创最新推荐文章于 2025-02-03 15:37:10 发布 · 219 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解专栏收录该内容

288 篇文章

订阅专栏

本文介绍了一种算法，该算法将单链表中的奇数节点和偶数节点进行分组，使得所有奇数节点位于偶数节点之前。此算法采用原地操作的方式，并确保了O(1)的空间复杂度和O(nodes)的时间复杂度。

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.

Nothing special.... set a flag to check whether the node is of even/odd number.

    ListNode* oddEvenList(ListNode* head) {
        if(!head || !head->next) return head;
        ListNode* oddDummy = new ListNode(0);
        ListNode* newHead = oddDummy;

        ListNode* evenDummy = new ListNode(0);
        ListNode* evenHead = evenDummy;

        ListNode* tmp = head;
        bool odd = true;
        while(tmp) {
            ListNode* next = tmp->next;
            if(odd) {
                oddDummy->next = tmp;
                oddDummy = tmp;
            } else {
                evenDummy->next  = tmp;
                evenDummy = tmp;
            }
            tmp = next;
            odd = !odd;
        }
        oddDummy->next = evenHead->next;
        evenDummy->next = NULL;
        return newHead->next;
    }