leetcode_25题——Reverse Nodes in k-Group （链表）

Reverse Nodes in k-Group

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Question Solution

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

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For 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

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这道题可以采用一种循环递归的方法来做，每次逆序k个节点，然后对于后面的由于和前面的一样，所以可以采用递归的方法

#include<iostream>
using namespace std;

struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

ListNode* reverseKGroup(ListNode* head, int k) {
	if(k<=1||head==NULL)
		return head;
	ListNode *ptr0=head;
	ListNode *ptr1=head;
	ListNode *head1=NULL;
	ListNode *ptr2=NULL;

	int N=k-1;
	int flag=0;
	while(N--)
	{
		ptr1=ptr1->next;
		if(ptr1==NULL)
		{
			flag=1;
			break;
		}
	}
	if(flag==1)
		return ptr0;
	ptr2=ptr1->next;

	ListNode *ptr3=ptr0->next;
	if(ptr3==ptr1)
	{
		ptr1->next=ptr0;
		ptr0->next=reverseKGroup(ptr2,k);
		return ptr1;
	}
	ListNode *ptr4=NULL;
	ptr0->next=reverseKGroup(ptr2,k);
	while(1)
	{
		ptr4=ptr3->next;
		ptr3->next=ptr0;
		if(ptr4==ptr1)
			break;
		else
		{
			ptr0=ptr3;
			ptr3=ptr4;
		}
	}
	ptr1->next=ptr3;
	return ptr1;
}

int main()
{

}