题解 | #链表中的节点每k个一组翻转#

```/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */

/**
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    struct ListNode *p1, *p2, *p3, *q, *p;
    p1 = pHead1; p2 = pHead2;
    p3 = NULL;
    struct ListNode *Tar;
    Tar = p3;
    if(p1 == NULL && p2 == NULL) return p1;
    if(p1 == NULL && p2 != NULL) return p2;
    if(p1 != NULL && p2 == NULL) return p1;
    while(p1 && p2) {
        if(p1->val >= p2->val) {
            if(Tar == NULL) {
                Tar = p2;
                p3 = p2;
                p2 = p2->next;
                p3->next = NULL;
                
            } else {
                q = p2;
                p2 = p2->next;
                q->next = Tar->next;
                Tar->next = q;
                Tar = q;
            }
        } else {
            if(Tar == NULL) {
                Tar = p1;
                p3 = p1;
                p1 = p1->next;
                p3->next = NULL;
                
            } else {
                q = p1;
                p1 = p1->next;
                q->next = Tar->next;
                Tar->next = q;
                Tar = q;
            }
        }
    }
    Tar->next = p2?p2:p1;
    return p3;
}