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 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.

递归:

/**
 * 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) {
        if (head == null || head.next == null || k < 2)
            return head;
        
        ListNode nextGroup = head;
        for (int i = 0; i < k; i++) {
            if (nextGroup != null)
                nextGroup = nextGroup.next;
            else
                return head;
        }
        
        ListNode nextGroupNew = reverseKGroup(nextGroup, k);
        ListNode prev = null, curr = head;
        while (curr != nextGroup) {
            ListNode next = curr.next;
            curr.next = prev != null ? prev : nextGroupNew;
            prev = curr;
            curr = next;
        }
        return prev;
    }
}

迭代:

/**
 * 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) {
        if (head == null || head.next == null || k < 2)
            return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode begin = dummy;
        int i = 0;
        
        while (head != null) {
            i++;
            if (i % k == 0) {
                begin = reverse(begin, head.next);
                head = begin.next;
            } else
                head = head.next;
        }
        return dummy.next;
    }
    
    private ListNode reverse(ListNode begin, ListNode end) {
        ListNode curr = begin.next, prev = begin;;
        ListNode next, first;
        first = curr;
        
        while (curr != end) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        begin.next = prev;
        first.next = curr;
        
        return first;
    }
}