[LeetCode]328.Odd Even Linked List | 奇偶链表操作

题目来源：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:
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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方案1：单一链表内指针操作，实现奇偶元素位置调整

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struct ListNode* oddEvenList(struct ListNode* head) {
    struct ListNode *p, *q, *m; 
    // Notice: head is the pointer which pointed to the first element of the list
    p = head;// p point to the last element of the odd ordered list
    if(p == NULL)
        return head;
    q = p->next;// q point to the last element of the even ordered list
    if(q == NULL)
        return head;
    while(p->next != NULL && q->next != NULL)
    {
        m = p->next;    
        p->next = q->next;
        q->next = q->next->next;
        p->next->next = m;
        p = p->next;
        q = q->next;
        if(p == NULL || q == NULL)
            return head;
    }
    return head;
}

这个思路挺有意思，画图说明下：
odd even lined list

方案2：两个链表头分别指向基数链表和偶数链表，最终连接奇偶链表

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ListNode* oddEvenList(ListNode* head) {
    if (!head) return head;
    ListNode* odd = head;
    ListNode* even = head->next;
    ListNode* evenHead = even;
    while (odd->next && even->next) {
        odd->next = even->next;
        odd = odd->next;
        even->next = odd->next;
        even = even->next;
    }
    odd->next = evenHead;
    return head;
}