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* p1 = head;
    struct ListNode* p2 = head->next;
    struct ListNode* eventHead = p2;
    while (p2 != NULL && p2->next != NULL) {
        p1->next = p2->next;
        p1 = p1->next;
        p2->next = p1->next;
        p2 = p2->next;
    }
    p1->next = eventHead;
    return head;
}