描述：

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.

解題思路：

將連結串列中的數值轉入到陣列vector中，對vector進行處理後再將資料按照順序傳給連結串列中的節點。

C++程式碼：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) 
    {
        vector<int>result;
        ListNode*ptemp = head;
        while
 
(ptemp!=NULL)
        {
            result.push_back(ptemp->val);
            ptemp = ptemp->next;
        }
        int count = result.size()/k;
        for(int i = 0;i<count;i++)
        {
            reverse(result.begin()+k*i,result.begin()+k*(i+1)); 
        }
        ListNode*temp = head;
        int i = 0;
        while(temp!=NULL)
        {
            temp->val= result[i];
            i++;
            temp = temp->next;
        }
        return head;
    }
};

執行結果：