25. K个一组翻转链表

25. K个一组翻转链表

遇到困难直接看题解  --_--,每k个一组，进行翻转，然后再装回原链表中。

class Solution {
       public ListNode reverseKGroup(ListNode head, int k) {
        ListNode hair = new ListNode(0); //头结点
        hair.next = head;
        ListNode pre = hair;

        while (head != null) {
            ListNode  tail = pre;
            //长度小于k时直接返回
            for (int i = 0 ; i < k; i++) {
                tail = tail.next;
                if (tail == null)
                    return hair.next;
            }
            ListNode nex = tail.next;
            ListNode[] reverse = myReverse(head,tail);
            head = reverse[0];
            tail = reverse[1];
            //子链表装回原链表
            pre.next = head;
            tail.next = nex;
            pre = tail;
            head = tail.next;
        }
        return hair.next;

    }
    public ListNode[] myReverse(ListNode head, ListNode tail) {
        ListNode prev = tail.next;
        ListNode p = head;
        while(prev != tail) {
            ListNode nex = p.next;
            p.next = prev;
            prev = p;
            p = nex;
        }
        return new ListNode[] {tail, head};
    }

}