leetcode 328 328. Odd Even Linked List 将链表的奇数偶数位置分开

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->NULLOutput: 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 ...

思路就是 新建两个结点一个表示偶数 一个表示奇数；然后复制过去就ok了  修改的时候 主要要备份一份；

CODE：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* oddEvenList(ListNode* head) 
    {
        if(head==nullptr)
            return nullptr;
        ListNode *dummy=new ListNode(0);
        dummy->next=head;
        ListNode *evenHead=head->next;
        ListNode *oddHead=head;
        ListNode *odd=oddHead;
        ListNode *even=evenHead;
        
        while(odd->next&&even->next)
        {
            odd->next=even->next;
            even->next=even->next->next;
            odd=odd->next;
            even=even->next;
        }
        odd->next=evenHead;//这里是把奇数偶数链表链接起来；所以要保存偶数链的起点位置；
        return dummy->next;

    }
};