Reverse Nodes in k-Group
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
分析
從頭節點到尾節點，依次翻轉K個節點，翻轉K個節點時，記錄K個翻轉後節點的頭節點指針和尾節點指針。

C++代碼

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int
 
 k) {
        ListNode*p,*q1=head,*lastNode=head,*q2,*last;
        if (!head||k==0)
            return head;
        q2 = nodeAfterKNodes(q1, k);
        head = reverseList(q1, k, last);
        lastNode = last;
        q1 = q2;
        while (q1){ 
            q2 = nodeAfterKNodes(q1, k);
            lastNode->next
 
= reverseList(q1, k, last);
            lastNode = last;
            q1 = q2;
        }
        return head;
    }
    ListNode* reverseList(ListNode* head,int k,ListNode*& last){
        if (!head || k > length(head)){
            last = head;
            return head;
        }   
        ListNode* p = head, *q=NULL, *r=NULL;
        last = head;
        p = head;
        q = head->next;
        head->next = NULL;
        if (q)
            r = q->next;
        int i = 0;
        while (q && i<k-1){
            q->next = p;
            p = q;
            q = r;
            if (r)
                r = r->next;
            i++;
        }
        return p;
    }
    int length(ListNode* head){
        int len = 0;
        while (head){
            len++;
            head = head->next;
        }
        return len;
    }
    ListNode* nodeAfterKNodes(ListNode* head, int k){
        if (k > length(head))
            return NULL;
        int i = 0;
        while (i < k){
            head = head->next;
            i++;
        }
        return head;
    }
};

leetcode:Reverse Nodes in k-Group