【leetcode】 Reverse Nodes in k-Group

地址

https://leetcode.com/problems/reverse-nodes-in-k-group/description/

题目

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.
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

思路

限制不让开大的内存，需要原地操作。一次翻转用递归解决即可。

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* p;
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode *ret,*q = head, *tmp;
        p = ret = new ListNode(1);
        while(q!=NULL)
        {
            int cnt=0;
            for(ListNode *i=q;i!=NULL&&cnt!=k;tmp=i=i->next,cnt++)
                ;
            if(cnt==k)
                dfs(q,k),q=tmp;
            else 
            {
                while(q!=NULL)
                {
                    p->next = q;
                    p=p->next;
                    q=q->next;
                }
                break;
            }
        }
        p->next = NULL;
        return ret->next;
    }
    void dfs(ListNode *x, int k)
    {
        if(k-1)
            dfs(x->next, k-1);
        p->next = x;
        p = p->next;
    }
};