leetcode Reverse Nodes in k-Group翻转链表K个一组

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list's nodes, only nodes itself may be changed.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* node = head;
        //第一个k个
        for(int i=0;i<k;i++){
            if(!node)
                return head;//不到k个长
            node = node->next;
        }
        //node是last的后一个
        ListNode* new_head = reverse(head,node);
        //关键递归
        head->next = reverseKGroup(node,k);
        return new_head;
    }

    //翻转链表 1 2 3
    ListNode* reverse(ListNode* head,ListNode* last){
        //last是翻转链表的后一个节点
        ListNode *pre=last, *cur=head;
        while(cur != last){
            ListNode* Next = cur->next;
            cur->next = pre;
            pre = cur;
            cur = Next;
        }
        return pre;
    }
};