25 Reverse Nodes in k-Group

Tag: Linked List

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.

如果链表的长度不是k的长度，那么最后多余的部分不用翻转。

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.

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 dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        while(true){
            head = reverse(head, k);
            if(head == null) 
                break;
        }
        return dummy.next;
    }
    
    // head -> n1 -> n2 ... nk -> nk+1
    // =>
    // head -> nk -> nk-1 .. n1 -> nk+1
    // return n1
    public ListNode reverse(ListNode head, int k){
        // 找到nk
        // 找到nkplus
        // 找到n1
        ListNode nk = head;
        for(int i = 0; i < k; i++){
            if(nk == null) return null; // 判断是否到了链表尾部不足k个元素的部分
            nk = nk.next;
        }
        
        if(nk == null) return null;   // 解决最后一个sub-linked-list长度不够的情况
        ListNode nkplus = nk.next;
        ListNode n1 = head.next;
        
        // 设置n1前面的node -> pre
        // 设置当前node -> cur
        // reverse the sub - linked list
        
        ListNode pre = null;
        ListNode cur = n1;
 
        while(cur != nkplus){
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        
        // connect
        head.next = nk;
        n1.next = nkplus;
        return n1; 
    }
}

Space Complexity

Space: O(1)