Java [leetcode 25]Reverse Nodes in k-Group

標籤：

題目描述：

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 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

解題思路：

仍然是採用遞迴的方法，跟上一題類似。按照k個數字為一段的方法遞迴判斷。如果該段長度不夠k，那麼直接返回該段的值。否則將該段k個數字從第一個數字開始兩個兩個反轉。

代碼如下：

public ListNode reverseKGroup(ListNode head, int k) {if (head == null)return null;ListNode tmp = head;for (int i = 1; i < k; i++) {tmp = tmp.next;if (tmp == null)return head;}ListNode nextRoundHead = tmp.next;ListNode firstNode = head;ListNode secondNode = head.next;ListNode thirdNode;for (int i = 1; i < k; i++) {thirdNode = secondNode.next;secondNode.next = firstNode;firstNode = secondNode;secondNode = thirdNode;}head.next = reverseKGroup(nextRoundHead, k);return tmp;}

 

Java [leetcode 25]Reverse Nodes in k-Group