题解 | #链表的奇偶重排#c语言迭代

struct ListNode* oddEvenList(struct ListNode* head ) {
    // write code here
    if(head==NULL||head->next==NULL||head->next->next==NULL)
        return head;//链表个数小于3直接返回
    struct ListNode *odd=head,*even=head->next;
    struct ListNode *Ehead=even;
    int oddnum=0;//链表长度为奇数标志位
    while(even->next!=NULL)
    {
        odd->next=even->next;
        odd=odd->next;
        if(odd->next==NULL)
        {
            oddnum=1;
            break;
        }    
        even->next=odd->next;
        even=even->next;
    }
    odd->next=Ehead;
    if(oddnum==1)
        even->next=NULL;
    return head;
}