【LeetCode】第17天 - 25. K 个一组翻转链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode newHead = new ListNode(0); //新链表的头
        newHead.next = head;
        ListNode tempHead = newHead;                // 子链表的头

        ListNode tail = newHead;                        //子链表的尾

        while(head != null){
            for(int i=0;i<k;i++){
                tail = tail.next;                   //head~tail表示一个子链表
                if(tail == null){                   //最后不足k个节点时直接返回
                    return newHead.next;
                }
            }
            ListNode next = tail.next;
            ListNode[] reverse = reverse(head, tail);   //翻转子链表
            head = reverse[0];
            tail = reverse[1];
            // 把子链表重新接回新链表
            tempHead.next = head;
            tail.next = next;
            tempHead = tail;
            head = tail.next;
        }
        return newHead.next;
    }
    //翻转head~tail的链表
    public ListNode[] reverse(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};
    }
}