328. Odd Even Linked List

Leetcode Linked List

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 $O(\text{nodes})$ 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 ...

分析

最直接的办法：一个odd链表保存奇数节点，一个even链表保存偶数节点。最后将odd链表的尾部只想even链表的首部，并将even链表的尾部指向null。

public ListNode oddEvenList(ListNode head) {
    if (head == null || head.next == null) return head;
    ListNode odd = head, even = head.next, cur = even.next;
    ListNode evenHead = even;
    // 一个odd链表保存奇数节点，一个even链表保存偶数节点
    while (cur != null && cur.next != null) {
        odd.next = cur;
        even.next = cur.next;
        // move to next 
        odd = odd.next;
        even = even.next;
        cur = cur.next.next;
    }

    // odd链表的尾部只想even链表的首部，并将even链表的尾部指向null。
    if (cur == null) {
        odd.next = evenHead;
        even.next = null;
    }
    else {
        //还有一个odd节点
        odd.next = cur;
        odd.next.next = evenHead;
        even.next = null;
    }
    return head;
}