[LeetCode] 25. Reverse Nodes in k-Group ☆☆☆

Posted 2020-08-31 Strugglion

 

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.

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

 

解法：

　　将链表分成k个一组，按顺序每次翻转一组。

　　对于每一组，先找到组内的第k个节点，如果找不到则说明长度小于k，无须翻转，返回null。

　　翻转时，先定位组内的头尾节点n1和nk。定义两个指针prev和curr，每次将curr.next变成prev后，两个指针分别后移一位，直到prev为nk，此时组内节点已全部翻转，只需再处理头尾节点即可。处理头尾节点，需要将n1.next指向curr（即nk＋1），再将head.next指向nk，最后返回n1。

/**
 * 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 (head == null || k < 2) {
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        
        while (head != null) {
            head = reverseK(head, k);
        }
        
        return dummy.next;
    }
    
    public ListNode reverseK(ListNode head, int k) {
        ListNode nk = head;
        for (int i = 0; i < k; i++) {
            nk = nk.next;
            if (nk == null) {
                return null;
            }
        }
        
        ListNode n1 = head.next;
        
        ListNode prev = null;
        ListNode curr = n1;
        while (prev != nk) {
            ListNode temp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = temp;
        }
        n1.next = curr;
        head.next = nk;
        return n1;
    }
}