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
給你一個鏈表以及一個k,將這個鏈表從頭指針開始每k個翻轉一下。
鏈表元素個數不是k的倍數，最后剩余的不用翻轉。
樣例
給出鏈表 1->2->3->4->5
k = 2, 返回 2->1->4->3->5
k = 3, 返回 3->2->1->4->5

分析

我們設計一個翻轉k的算法
head -> n1 -> n2 ... nk -> nk+1 => head -> nk -> nk-1 .. n1 -> nk+1
return n1
那么我們只要接著對這個方法傳入n1即可，重復調用這個方法，直到遇到null

代碼

/**
 * 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) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        head = dummy;
        while (true) {
            head = reverseK(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 reverseK(ListNode head, int k) {
        ListNode nk = head;
        for (int i = 0; i < k; i++) {
            if (nk == null) {
                return null;
            }
            nk = nk.next;
        }
        
        if (nk == null) {
            return null;
        }
        
        // reverse        
        ListNode n1 = head.next;
        ListNode nkplus = nk.next; 
        
        ListNode prev = null;
        ListNode curt = n1;
        while (curt != nkplus) {
            ListNode temp = curt.next;
            curt.next = prev;
            prev = curt;
            curt = temp;
        }
        
        // connect
        head.next = nk;
        n1.next = nkplus;
        return n1;
    }
}