[leetcode]25. Reverse Nodes in k-Group每k个节点反转一下

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.

题意：

给定一个链表，每k个节点一组，做一次反转。

Solution1：we can still simplify this question into how to reverse a linked list, the only difference is we need to set "dummy" and "null" like the left and right boundary by ourselves.

(1) set a pointer pre as a " dummy " ahead

(2) set a pointer last as a " null " boundary

(3) iteratively move cur into the front(pre.next)

(4) cur = next 

(5)iteratively move cur into the front(pre.next) until cur meet the "null" boundary

code:

/*
Time: O(n)
Space: O(1)
*/
class Solution {
     public ListNode reverseKGroup(ListNode head, int k) {
            if (head == null) return null;
            ListNode dummy = new ListNode(-1);
            ListNode pre = dummy;
            dummy.next = head;
            // to reverse each k-Group, considering pre as a "dummy" ahead
            while (pre != null) {
                pre = reverse(pre, k);
            }
            return dummy.next;
        }
    
        public ListNode reverse(ListNode pre, int k) {
            // to reverse each k-Group, considering last as a "null"  boundary
            ListNode last = pre;
            for (int i = 0; i < k + 1; i++) {
                last = last.next;
                if (i != k && last == null) return null;
            }
            
            // reverse 
            ListNode tail = pre.next;
            ListNode cur = pre.next.next;
            // remove cur to front, then update cur
            while (cur != last) {
                ListNode next = cur.next;
                cur.next = pre.next;
                pre.next = cur;
                tail.next = next;
                cur = next;  
            }
            return tail;
        }
}