LinkedList - 328. Odd Even Linked List

328. Odd Even 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(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 ...

思路：

题目意思将链表中的奇数位节点放在链表前面，偶数位节点放在链表后面，简单的实现题，注意的点主要是不要产生断链（记录号偶数节点的起始位置）和循环结束条件，避免出现空针等情况。

代码：

golang:

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func oddEvenList(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    
    var oddHead, odd, evenHead, even *ListNode
    odd = head
    even = head.Next
    oddHead = odd
    evenHead = even
    
    for even != nil && even.Next != nil {
        odd.Next = even.Next
        odd = odd.Next
        
        even.Next = odd.Next
        even = even.Next
        
    }
    odd.Next = evenHead
    
    return oddHead
}

java：

/**

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