LeeCode 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:
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 ...

Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

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  题目：给你一个单向链表，每个奇数后面都会跟一个偶数，请你把奇数全部移到一起，后面跟的全部是偶数，而且他们原来的相对次序不变，第一个点认为是奇数，第二个点认为是偶数。

 

这明显是一个单向链表的移动问题，要求的时间复杂度为O(nodes)空间复杂度为O(1)。我们做的是，把后面的奇数删掉，并把这个点移到前面来，然后向后移动奇数的标记点一位，偶数的标记点一位，最后的奇数点指向第一个偶数点即可，代码如下：

/**
 * 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==NULL)
        return head;
        ListNode *beg=head,*end=head->next,*temp=end;
        while(end!=NULL&&end->next!=NULL)
        {
            beg->next=end->next;
            end->next=end->next->next;
            beg=beg->next;
            end=end->next;
        }
        beg->next=temp;
        return head;
    }
};