25. K 个一组翻转链表

迭代法

/**
 * 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 dummy = new ListNode(0);
        dummy.next = head;
        ListNode p = dummy;
        while (true) {
            ListNode q = p;
            // 统计是否存在链表后面是否存在k个节点
            for (int i = 0; i < k && q != null; i++) {
                q = q.next;
            }
            // 如果不足K个，则跳出循环
            if (q == null) {
                break;
            }
            ListNode a = p.next, b = a.next;
            // 这k个节点内部反转
            for (int i = 0; i < k - 1; i++) {
                ListNode c = b.next;
                b.next = a;
                a = b;
                b = c;
            }
            // 外部反转拼接
            ListNode c = p.next;
            p.next = a;
            c.next = b;
            p = c;
        }
        return dummy.next;
    }
}

• 时间复杂度 O(n)
• 空间复杂度 O(1)

递归法

/**
 * 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;
        for (int i = 0; i < k; i++) {
            // 如果不足k个节点，则不需要反转
            if (tail == null) {
                return head;
            }
            tail = tail.next;
        }
        // 反转前k个元素
        ListNode newHead = reverse(head, tail);
        // 下一轮的开始的地方就是tail
        head.next = reverseKGroup(tail, k);
        return newHead;
    }

    // 链表反转函数
    private ListNode reverse(ListNode head, ListNode tail) {
        ListNode prev = null;
        ListNode next = null;
        while (head != tail) {
            next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
}

• 时间复杂度 O(n)
• 空间复杂度 O(n / k)，递归所需空间