LeetCode第25題之Reverse Nodes in k-Group

基本思想是對每k個節點採用頭插法，時間複雜度爲O（n）。
C++代碼：

#include <vector>
#include <iostream>
using namespace std;

/**
 * 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) {
         if (k<2 || NULL == head || NULL == head->next)
             return head;
         //res->next記錄返回結果
         ListNode * res = new ListNode(-1);
         res->next = head;
         //last_tail記錄上一次對k個節點進行操作後的最後一個節點
         ListNode * last_tail = res;
         while(1)
         {
             //找t_head之後是否有k個節點
            int n = 0;
    //      ListNode *num_p = t_head;
            ListNode *num_p = last_tail->next;
            while(num_p && n != k)
            {
                ++n;
                num_p = num_p->next;
            }
            //如果n等於k，則num_p爲下一組的首節點的指針
            //不足k個節點，返回結果
            if(n != k)
            {
                ListNode * t = res->next;
                delete res;
                return t;
            }
            //n == k
            else
            {
                //pre記錄待插入節點的上一個節點
                ListNode *pre = last_tail->next;
                //cur記錄當前待插入的節點
                ListNode *cur = pre->next;
                //此時num_p爲以t_head開頭的第k+1個節點
                //將當前待插入的節點插到last_tail的後面
                while (cur != num_p)
                {
                    pre->next = cur->next;
                    cur->next = last_tail->next;
                    last_tail->next = cur;

                    //指針後移
                    cur = pre->next;
                }
                //k個節點插入完畢後，更新last_tail到當前操作的k個節點的最後一個節點
                last_tail = pre;
            }
         }
     }
 };
int main()
{
    Solution s;
    ListNode *head, *p;
    head = NULL;
    for (int i=7;i>0;--i)
    {
        p = new ListNode(i);
        p->next = head;
        head = p;
    }
    ListNode *res = s.reverseKGroup(head, 3);
    p = res;
    while(p)
    {
        cout<<p->val<<'\t';
        p = p->next;
    }
    return 0;
}