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 || head->next == NULL) return head;
    struct ListNode* p = head;
    struct ListNode* run1 = head->next;
    struct ListNode* is2 = head->next;
    while (run1 && run1->next) {
        p->next = run1->next;
        p = p->next;
        run1->next = p->next;
        run1 = run1->next;
    }
    p->next = is2;
    return head;
}