【Leetcode】之Reverse Nodes in k-Group

一.问题描述

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

二.我的解题思路

链表的问题都很直观。对于这道题目，很自然的想到分治递归的办法。以k个节点为一个单位，进行reverse求解。测试通过的程序如下：

/** * 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) {            ListNode* curr,* next,* tmp;            ListNode* st,* end;            int cnt=0;            curr=head;            if(curr==NULL) return NULL;             while(curr!=NULL){                cnt++;                if(cnt==k)                    break;                curr=curr->next;            }   //judge if there exists k nodes            if(cnt==k){ //if exists               st=head;cnt=1;end=head;                while(cnt<k){                    next=end->next;                    tmp=next->next;                    next->next=st;                    end->next=tmp;                    st=next;                    cnt++;                }                if(end->next==NULL)                    return st;                end->next=reverseKGroup(end->next,k);                return st;            }                        return head;                                                                                }};