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
     */
    void reverse(ListNode* head, int n, ListNode* end) {
        ListNode* pre = nullptr;
        ListNode* cur = head;
        for(int i=0; i<n; i++) {
            ListNode* tmp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = tmp;
        }
    }

    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* p = dummy;
        ListNode* pre = p;
        ListNode* leftNode = p;
        ListNode* rightNode = p;
        ListNode* suc = p;


        while(p != nullptr) {
            pre = leftNode;
            leftNode = pre->next;
            rightNode = pre;
            p = rightNode;
            for(int i=0; i<k; i++) {
                rightNode = rightNode->next;
                p = p->next;
                if(p == nullptr) {
                    return dummy->next;
                }
            }
            suc = rightNode->next;
            pre->next = nullptr;
            rightNode->next = nullptr;
            reverse(leftNode, k,rightNode);
            pre->next = rightNode;
            leftNode->next = suc;           
        }

        return dummy->next;

    }
};

我觉得需要画张图来进行判断每个指针的位置