25. Reverse Nodes in k-Group(need constant approach)

Description

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 of k then left-out nodes in the end should remain as it is.

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

Note:

Only constant extra memory is allowed.
You may not alter the values in the list’s nodes, only nodes itself may be changed.

Problem URL

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Solution

将一个链表以k个为一组，进行反转。

A recursive approach. Use a current node to move forward k steps. If it is valid, that means count == k, use cur as head node to recursively reverse remaining nodes. Then reverse node by putting head to previous of cur. After all of that, move head to cur to return the same item as not enough k circumstance.

Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode cur = head;
        int count = 0;
        //move cur k nodes further
        while (cur != null && count != k){
            cur = cur.next;
            count++;
        }
        if (count == k){
            //reverse remaining nodes recursively
            cur = reverseKGroup(cur, k);
            //move k nodes from head to prior of cur
            while(count-- > 0){
                ListNode temp = head.next;
                head.next = cur;
                cur = head;
                head = temp;
            }
            head = cur;
        }
        return head;
    }
}

Time Complexity: O(n)
Space Complexity: O(1)

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