题解 | #链表中的节点每k个一组翻转#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 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
     */

     ListNode* signlereverse(ListNode* head, int m, int n)
     {
         if(m == n)
            return head;

        ListNode* s1, *s2;
        
        int sn1 = 0, tmp;

        s1 = head;
        while(sn1++ < m - 1)
            s1 = s1->next;

        s2 = s1;
        for(int i = 0; i < n - m; i++){
            s1 = s2;
            for(int j = i; j < n - m; j++){
                tmp = s1->val;
                s1->val = s1->next->val;
                s1->next->val = tmp;
                s1 = s1->next;
            }
        }

         return head;       
     }

    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        int count = 1, tmp;
        ListNode *s1;

        if(!head)
            return head;

        s1 = head;
        while((s1 = s1->next) != nullptr)
            count++;

        count = count / k ;
        for(int i = 0; i < count; i++)
            signlereverse(head, i * k + 1, i * k + k);

        return head;
    }
};