题解 | #链表中的节点每k个一组翻转#

对于链表，我感觉我悟了。。。

 * 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
 */
/**
 * 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 
 * @param k int整型 
 * @return ListNode类
 */
struct ListNode* reverseKGroup(struct ListNode* head, int k ) {
    // write code here
    
    int n = 0;   //节点数
    //判断在哪个节点翻转
    if (head == NULL || head->next == NULL)
    {
        return head;
    }
    struct ListNode* pn; //判读链表的节点数
    pn = head;
    while (pn != NULL)
    {
        pn = pn->next;
        n = n+1;      //节点数为n
    }
    if (n<k)     //节点数小于k
    {
        return head;
    }
    
    struct ListNode* p0;
    struct ListNode* p1;    //定义需要连接的首尾指针
    struct ListNode* p2;
    p0 = NULL;
    p1 = NULL;
    p2 = NULL;
    //反转函数：
    struct ListNode* pre;
    struct ListNode* cur;
    struct ListNode* nex;
    cur = head;
    nex = cur->next;
    pre = cur;
    cur = nex;
    for (int j=1;j<=(n/k);j++)
    {
        p0 = pre;
        for(int i =1;i<k;i++)          //做反转
        {
            nex = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nex;
        }
        p1 = pre;
        if (j==1)
        {
            head = p1;
        }
        if (j>=2)
        {
            p2->next = p1;   //首尾相连
            p0->next = NULL;
        }
        p2 = p0;
        nex = cur->next;
        pre = cur;
        cur = nex;
    }
    p2->next = pre;
    return head;
}