25. K 个一组翻转链表

题目分析

按递归的方法，把整个链表划分为k个节点一组去处理，翻转前k个节点后，再对k+1节点后的链表进行递归处理，依次翻转链表。过程可以参考「206. 反转链表」

递归是一种算法结构。递归会出现在子程序中，形式上表现为直接或间接的自己调用自己

/**
 * 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) {
		if (head == null || head.next == null) {
			return head;
		}

		ListNode tail = head;
		// 剩余节点数不足k个直接返回
		for (int i = 0; i < k; i++) {
			if (tail == null) {
				return head;
			}
			tail = tail.next;
		}
		ListNode newHead = reverse(head, tail);
		head.next = reverseKGroup(tail, k);
		return newHead;
    }

	/**
	 * 翻转链表，指针翻转后，最终返回尾节点作为新链表的头结点
	 * @param head
	 * @param tail
	 * @return
	 */
	private ListNode reverse(ListNode head, ListNode tail) {
		ListNode pre = null, next = null;
		while (head != tail){
			next = head.next;
			head.next = pre;
			pre = head;
			head = next;
		}
		return pre;
	}
}