LeetCode -- Reverse Nodes in k-Group

LeetCode -- Reverse Nodes in k-Group
Description:


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 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


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


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


If the number of nodes from the current node is greater than k in the Process of the traversal chain table, the k nodes are reversed, and then k locations are moved backward from the current position, continue to reverse until the number of nodes from the current node to the end is less than K.


Ideas:
1. Create a reversal function to reverse the k nodes from the current start. Space complexity of this function O (n)
2. Get the total length of the linked list first. If it is smaller than k, return directly. Otherwise, k nodes are rotated each time and pointed to the position of the next rotation.




Implementation Code:

/** * Definition for singly-linked list. * public class ListNode { *     public int val; *     public ListNode next; *     public ListNode(int x) { val = x; } * } */public class Solution {    public ListNode ReverseKGroup(ListNode head, int k)     {        if(head == null || head.next == null || k == 1){    return head;    }        var len = 0;    var h = head;    while(head != null){    head = head.next;    len ++;    }        if(k > len){    return h;    }        ListNode pre = null;    ListNode newHead = null;    while(len >= k)    {    ReverseKNodes(ref h, k, ref pre);        if(pre == null){    newHead = h;    pre = h;    for(var i =0 ;i < k-1; i++){    pre = pre.next;    }    }else{    for(var i = 0;i < k;i++){    pre = pre.next;    }    }        for(var i = 0;i < k;i++){    h = h.next;    }        len -= k;    }        return newHead;    }private void ReverseKNodes(ref ListNode n, int k, ref ListNode preNode){var stack = new Stack
 
  ();for(var i = 0;i < k; i++){stack.Push(n.val);n = n.next;}ListNode end = n;ListNode tmp = new ListNode(stack.Pop());var tmpHead = tmp;while(stack.Count > 0){var n1 = new ListNode(stack.Pop());tmp.next = n1;tmp = tmp.next;}tmp.next = end;if(preNode != null){n = tmpHead;preNode.next = tmpHead;}else{n = tmpHead;}}}