【LeetCode刷题-链表】--25.K个一组翻转链表

25.K个一组翻转链表

思路：

把链表节点按照k个一组分组，可以使用一个指针head依次指向每组的头节点，这个指针每次向前移动k步，直至链表结尾，对于每个分组， 先判断它的长度是否大于等于k，若是，就翻转这部分链表，否则不需要翻转

/**
 * 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 hair = new ListNode(0);
        hair.next = head;
        ListNode pre = hair;
        while(head != null){
            ListNode tail = pre;
            //查看剩余部分长度是否大于等于k
            for(int i = 0 ;i<k;i++){
                tail = tail.next;
                if(tail == null){
                    return hair.next;
                }
            }
            ListNode next = tail.next;
            ListNode[] reverse = Reverse(head,tail);
            head = reverse[0];
            tail = reverse[1];
            //把子链表重新接回原链表
            pre.next = head;
            tail.next = next;
            pre = tail;
            head = tail.next;
        }
        return hair.next;

    }
    //反转链表，返回子链表的头部与尾部
    public ListNode[] Reverse(ListNode head,ListNode tail){
        ListNode prev = tail.next;
        ListNode p = head;
        while(prev != tail){
            ListNode next = p.next;
            p.next = prev;
            prev = p;
            p = next;
        }
        return new ListNode[]{tail,head};
    }
}
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