Odd Even Linked List

5-13 数据结构与算法 lynsa 1,658 views

Odd Even Linked List

5-13 1,658 views

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 …

给定一个单向链表，将奇数节点放在前面，偶数节点放在后面。所有奇数节点或偶数节点的相对位置保持不变。空间复杂度O(1)，时间复杂度O(n)。所有奇数节点或偶数节点的相对位置保持不变。

/**
 * 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) {
        ListNode n1 = head;
        if (n1 == null) return head;
        ListNode head2 = head.next;
        ListNode n2 = head.next;
        while (n2 != null && n2.next != null) {
            n1.next = n2.next;
            n1 = n2.next;
            n2.next = n1.next;
            n2 = n1.next;
        }
        n1.next = head2;
        return head;
    }
}

/**

* 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) {

ListNode n1 = head;

if (n1 == null) return head;

ListNode head2 = head.next;

ListNode n2 = head.next;

while (n2 != null && n2.next != null) {

n1.next = n2.next;

n1 = n2.next;

n2.next = n1.next;

n2 = n1.next;

}

n1.next = head2;

return head;

}

版权属于: 我爱我家

原文地址: http://magicwt.com/2014/05/13/odd-even-linked-list/

转载时必须以链接形式注明原始出处及本声明。