【Leetcode】：328. Odd Even Linked List 问题 in JAVA

永不放弃EX

于 2016-04-28 21:44:12 发布

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分类专栏：编程 JAVA 文章标签： leetcode java

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本文介绍了一种在O(1)空间复杂度和O(n)时间复杂度下，将单链表中的奇数位置节点与偶数位置节点进行重组的方法。通过遍历链表并调整节点指针的方式，确保了奇数位置节点位于前面，而偶数位置节点位于其后，同时保持了原有的相对顺序。

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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 ...

题目难度是中等，要求将所有偶数编号的节点放在奇数编号的节点后面，空间复杂度要求为O（1）

/**
 * 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) {
            return null;
        }
        ListNode node = head;
        ListNode even = head.next;
        while(node.next != null && node.next.next != null) {
            ListNode next = node.next;
            if (next.next != null) {
                node.next = next.next;    
                next.next = next.next.next;
                node = node.next;         
            } else {
                node.next = next.next;
                node = node.next;
            }
        }
        if (even != null) {
            node.next = even;
        }
        return head;
    }
}