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* reverseKGroup(ListNode* head, int k) {
        // write code here
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* p = head;
        int n = 0;
        while(p) {
            p=p->next;
            n++;
        }
        ListNode* p0=dummy;
        ListNode* cur = p0->next;
        ListNode* pre = nullptr;
        ListNode* next = nullptr;
        while(n>=k) {
            n-=k;
            for(int i=1;i<=k;i++) {
                next = cur->next;
                cur->next = pre;
                pre = cur;
                cur = next;
            }
            next = p0->next;
            p0->next->next = cur;
            p0->next = pre;
            p0=next;
        }
        return dummy->next;
    }
};