[Leetcode click Notes] Reverse nodes in K-group

Given a linked list, reverse the nodes of a linked listKAt a time and return its modified list.

If the number of nodes is not a multipleKThen left-out nodes in the end shoshould 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

ForK= 2, you shoshould return:2->1->4->3->5

ForK= 3, you shoshould return:3->2->1->4->5

 

Question: The idea is simple. You can do it step by step. In addition to the functions required by the question, two functions are implemented:

Private listnode [] reversesub (listnode head, int K) This function reverts the chain table whose head points to K and returns an array of 2, the first element in the array is the head node of the reverse linked list, and the second element is the end node of the reverse linked list.

Private int getlength (listnode head) This function returns the length of the chain table pointed to by the head.

In the reversekgroup function, first calculate the length Len of the original linked list, then the number of groups to be reversed is Len/K, and then call the reversesub function Len/K to reverse each chain table, then string them based on the returned first and last pointers. Finally, judge whether there are elements in the linked list that do not need to be reversed based on whether Len % K is 0. If so, link them to the end of the linked list and return them.

The Code is as follows:

 1 /** 2  * Definition for singly-linked list. 3  * public class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { 7  *         val = x; 8  *         next = null; 9  *     }10  * }11  */12 public class Solution {13     private ListNode[] reverseSub(ListNode head,int k){14         ListNode[] answer = new ListNode[2];15         answer[1] = head;16         ListNode prev = null;17         for(int i = 0;i < k;i++){18             ListNode temp = head.next;19             head.next = prev;20             prev = head;21             head = temp;22         }23         answer[0] = prev;24         return answer;25     }26     private int getlength(ListNode head){27         int count = 0;28         while(head != null){29             count ++;30             head = head.next;31         }32         return count;33     }34 35     public ListNode reverseKGroup(ListNode head, int k) {36         int len = getlength(head);37         if(len < k)38             return head;39         40         ListNode answer = null;41         ListNode tail = new ListNode(0);42         int for_num = len / k;43         for(int i = 0;i < for_num;i++){44             ListNode h = head;45             46             //find next starter47             for(int j = 0;j < k;j++)48                 head = head.next;49             50             ListNode[] temp = reverseSub(h, k);51             if(answer == null){52                 answer = temp[0];53                 tail = temp[1];54             }55             else {56                 tail.next = temp[0];57                 tail = temp[1];58             }                        59         }60         61         if(len%k != 0)62             tail.next = head;63         else64             tail.next = null;65         66         return answer;67     }68 }