25. 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.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple ofk 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

Solution:

Tips:

Just use two pointers to record a group of elements.

Java Code :

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode reverseKGroup(ListNode head, int k) {        if (null == head || k == 1) {            return head;        }                ListNode newHead = new ListNode(0);        ListNode groupTail = newHead; // record the last element of the group        // pointer1 close to head        ListNode h = head;        // pointer2 k steps after h        ListNode t = head;        int n = k;        // find a group of k elements        while (t != null && n > 0) {            t = t.next;            n--;        }        // list has elements less than k        if (n > 0) {            return head;        }                while (h != null && n == 0) {            n = k;            ListNode gt = h;            // reverse group elements            while (n > 0) {                ListNode p = h;                h = h.next;                                p.next = groupTail.next;                groupTail.next = p;                n--;            }                        // forward k steps            n = k;            while (t != null && n > 0) {                t = t.next;                n--;            }                        groupTail = gt;        }                if (n > 0) {            groupTail.next = h;        }                return newHead.next;    }}