【LeetCode & 剑指offer刷题】链表题10：328 Odd Even Linked List

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

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.（相对顺序要保持）
• The first node is considered odd（odd为奇数，相当于1序开始）, the second node even and so on ...

 

/*

问题：在链表中，将所有奇数序号的结点放到前面，偶数序号的结点放在后面，要求就地解决

与问题“调整数组中奇数偶数顺序”区别在于前者调整结点，而后者调整的是值

O(n),O(1)

*/

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

 *     ListNode(int x) : val(x), next(NULL) {}

 * };

 */

//方法：head、evenhead.两个链表，再拼在一起

class Solution

{

public:

    ListNode* oddEvenList(ListNode* head)

    {

        if(!head) return head;

        ListNode* odd = head; //奇数序列结点指针与头指针

        ListNode* evenhead = head->next,*even = evenhead; //偶数序列结点指针与头指针

       

        while(even && even->next) //偶数序列指针判断(循环时一般用后面的指针来判断是否结束循环) 对于走两步的指针p均需要判断p与p->next是否为空

        {

 

            odd->next = odd->next->next; //连接奇数序列结点,每次走两步 ，先连接前面的指针

            even->next = even->next->next;//连接偶数序列结点

           

            odd = odd->next; //指向下一个奇结点

            even = even->next;

        }

        odd->next = evenhead; //连接奇序列链表和偶序列链表

        return head;

       

    }

};