328. Odd Even Linked List

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本文介绍了一种链表操作技巧，将链表中的奇数位置节点与偶数位置节点进行分离并重组，同时保持各自相对顺序不变。通过两种方法实现，确保空间复杂度为O(1)，时间复杂度为O(nodes)。

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题目描述

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 …

方法思路

Approach1:

class Solution {
    //Runtime: 3 ms, faster than 99.15% 
    //Memory Usage: 39.7 MB, less than 5.17%
    public ListNode oddEvenList(ListNode head) {
        ListNode odd = new ListNode(0), oddHelp = odd;
        ListNode even = new ListNode(0), evenHelp = even;
        while(head != null && head.next != null){
            oddHelp.next = head;
            oddHelp = oddHelp.next;
            head = head.next;
            evenHelp.next = head;
            evenHelp = evenHelp.next;
            head = head.next;
        }
        if(head == null)
            oddHelp.next = null;
        else if(head.next == null){
            oddHelp.next = head;
            evenHelp.next = null;
        }
        
        return merge(odd.next, even.next);
    }
    
    public ListNode merge(ListNode l1, ListNode l2){
        if(l1 == null) return l2;
        else if(l2 == null) return l1;
        ListNode help = l1;
        while(help.next != null) help = help.next;
        help.next = l2;
        return l1;
    }
}

Apporach2:

class Solution{
    //Runtime: 3 ms, faster than 99.15% of Java
    //Memory Usage: 37.3 MB, less than 31.93% of Java 
    public ListNode oddEvenList(ListNode head){
        if(head == null || head.next == null)
            return head;
        
        ListNode odd = head, even = head.next, evenHead = even;
        while(even != null && even.next != null){
            odd.next = even.next;
            odd = odd.next;
            even.next = odd.next;
            even = even.next;
        }
        
        odd.next = evenHead;
        return head;
    }
}