题解 | #链表中的节点每k个一组翻转#

import java.util.*;
public class Solution {
    public ListNode reverseKGroup (ListNode head, int k) {
        // write code here
        if (head == null)
            return head;
        ListNode head1 = head;
        //记录链表总长度
        int length = 1;
        while(head1.next!=null){
            head1 = head1.next;
            length++;
        }
        // System.out.print("length="+length);
        head1 = head;
        int arr [] = new int [length];
        int i =0;
        //将链表中的值放入数组中
        while(i<length){
            arr[i++] = head1.val;
            head1 = head1.next;
        }
        head1 = head;








        //     System.out.print("first:");
        // for(int j :arr){
        //     System.out.print(j);
        // }
        //count指的是反转count次
        int count = length/k;
        //count1指的是每次反转调换count次
        int count1=k/2;     
        // System.out.println("count = "+count+"count1="+count1);
        for(int n = 0;n<count;n++){
            // 反转count次，每次循环，反转的部分总长度为n*k-n*k+k，基础长度为n*k
		  	//所以调换的元素为n*k和n*k+k，在m1的大小中调换
                for(int m1 =0;m1<count1;m1++){
                int temp = arr[n*k+m1];
                arr[n*k+m1] = arr[n*k+k-m1-1];
                arr[n*k+k-m1-1] = temp;
            }
        }
            // System.out.print("secound:");
        for(int j :arr){
            // System.out.print(j);
            head1.val = j;
            head1 = head1.next;
        }
        return head;
    }
}