leetcode_25_Reverse Nodes in k-Group

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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

class Solution {//once the link order of the list is changed, //we must keep the previous pointer(p in the code) updated, if we want to use it agianpublic:void reverse_range(ListNode* prev, ListNode* end, ListNode*& p){ListNode* last = prev->next;ListNode* cur = last->next;while(cur != end){last->next = cur->next;cur->next = prev->next;prev->next = cur;cur = last->next;}p = last;}ListNode *reverseKGroup(ListNode *head, int k) {if(k <= 1) return head;ListNode *dummy = new ListNode(-1);dummy->next = head;ListNode* prev = dummy;ListNode* p = dummy->next;int cnt = 1;while(p != NULL){if(cnt%k == 0){reverse_range(prev, p->next, p);//update p is very importantprev = p;}p = p->next;cnt++;}return dummy->next;}};

#include<iostream>using namespace std;#define N 5struct ListNode {int val;ListNode *next;ListNode(int x) : val(x), next(NULL) {}};class Solution {//once the link order of the list is changed, //we must keep the previous pointer(p in the code) updated, if we want to use it agianpublic:void reverse_range(ListNode* prev, ListNode* end, ListNode*& p){ListNode* last = prev->next;ListNode* cur = last->next;while(cur != end){last->next = cur->next;cur->next = prev->next;prev->next = cur;cur = last->next;}p = last;}ListNode *reverseKGroup(ListNode *head, int k) {if(k <= 1) return head;ListNode *dummy = new ListNode(-1);dummy->next = head;ListNode* prev = dummy;ListNode* p = dummy->next;int cnt = 1;while(p != NULL){if(cnt%k == 0){reverse_range(prev, p->next, p);//update p is very importantprev = p;}p = p->next;cnt++;}return dummy->next;}};ListNode *creatlist(){ListNode *head = NULL;ListNode *p;for(int i=0; i<N; i++){int a;cin>>a;p = (ListNode*) malloc(sizeof(ListNode));p->val = a;p->next = head;head = p;}return head;}int main(){ListNode *list = creatlist();Solution lin;int a;cin>>a;ListNode *outlist = lin.reverseKGroup ( list,a );for(int i=0; i<N; i++){cout<<outlist->val;outlist = outlist->next;}}