题解 | #链表的奇偶重排#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param head ListNode类
 * @return ListNode类
 */
struct ListNode* oddEvenList(struct ListNode* head ) {
    // write code here
    if(head==NULL)
    return head;
    struct ListNode* head1 = NULL;
    struct ListNode* ptail1 = NULL;
    struct ListNode* head2 = NULL;
    struct ListNode* ptail2 = NULL;
    struct ListNode* p1 = head, *p2 = head->next;
    while (p1) {
        if (head1 == NULL)
            head1 = ptail1 = p1;
        else {
            ptail1->next = p1;
            ptail1 = p1;
        }
        if (head2 == NULL)
            head2 = ptail2 = p2;
        else {
            ptail2->next = p2;
            ptail2 = p2;
        }
        if(p2 == NULL)
        break;
        p1 = p1->next->next;
        if (p2->next != NULL)
            p2 = p2->next->next;
    }
    ptail1->next = head2;
    return head;
}