题解 | #链表中的节点每k个一组翻转#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    
    ListNode* reverse(ListNode* head){
        ListNode* cur,*pre,*next;
        pre = nullptr;
        cur = head;
        next = cur->next;
        while(next){
            cur->next = pre;
            pre = cur;
            cur = next;
            next = next->next;
        }
        cur->next = pre;
        return cur;
    }
    
    int get_size(ListNode* head){
        ListNode* temp = head;
        int size=0;
        while(temp){
            size++;
            temp = temp->next;
        }
        return size;
    }
    
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) {
        int cnt = 0;
        int size=0;
        size = get_size(head);
        if(size==0) return nullptr;
        if(size==1||k==1||size<k) return head;
        ListNode *p1,*p2,*p3,*p4, *ret,*cur,*temp;
        cur = head;
        ret = nullptr;
        p1 = nullptr;
        p2 = head;
        //p3 = head;
        while(cur){
            if(++cnt==k){
                cnt=0;
                p3 = cur;
                p4 = p3->next;
                p3->next = nullptr;
                temp = reverse(p2);
                if(ret==nullptr) ret = p3;
                else p1->next = temp;
                p1 = p2;
                p2 = p4;
                cur = p4;
                continue;
                
                
            }
            cur = cur->next;
        }
        if(cnt<k) p1->next = p4;
        return ret;
    }
};