Odd Even Linked List

最新推荐文章于 2022-08-07 09:55:51 发布

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最新推荐文章于 2022-08-07 09:55:51 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/sinat_29621543/article/details/53501709

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本文介绍了一种链表操作算法，该算法能在O(1)的空间复杂度和O(nodes)的时间复杂度下，将链表中的奇数位置节点与偶数位置节点重新排列。通过创建两个子链表，一个由所有奇数位置节点组成，另一个由偶数位置节点组成，最终将奇数节点链表的末尾指向偶数节点链表的头结点。

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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:
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 ...

题目意思是将链表奇数位的节点排在偶数位前面，解法就是产生两条链表，一条都是奇数位的链表head,一条偶数位节点组成的链表evenHead,最后将head的末位节点指向evenHead

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }
        ListNode even = head.next;
        ListNode evenHead = head.next;
        ListNode tail = head;
        while(even != null){
            tail.next = even.next;
            if(even.next != null){
                even.next = even.next.next;
                tail = tail.next;
            }
            even = even.next;
        }
        tail.next = evenHead;
        return head;
    }
}