题解 | #链表中的节点每k个一组翻转#

//1首先要明确翻转多少次 1 2 3 4 5 k=2 times=5/2=2

//2确立一个答案链表 ans

//3建立一个栈

//4每次放k个元素进栈

//5每次出k个元素进答案链表 把4，5进行times次

//把剩下的不足以k个的接在答案链表

import java.util.*;
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        int length = 0;
        ListNode f = head;
        while(f!=null){
            length++;
            f=f.next;
        }
        int times = length/k;
        ListNode ans=new ListNode(-1);
        ListNode now = head;
        ListNode newone = ans;
        Stack<ListNode> stack = new Stack<>();
        for(int i = 0;i<times;i++) {
            for (int j = 0; j < k; j++) {
                stack.push(now);
                now = now.next;
            }
            for (int g = 0; g < k; g++) {
                newone.next = stack.pop();
                newone = newone.next;
            }
        }
        newone.next = now;
        return ans.next;
    }
}