0025. Reverse Nodes in k-Group

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Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list's nodes, only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Example 3:

Input: head = [1,2,3,4,5], k = 1
Output: [1,2,3,4,5]

Example 4:

Input: head = [1], k = 1
Output: [1]

Constraints:

• The number of nodes in the list is in the range sz.
• 1 <= sz <= 5000
• 0 <= Node.val <= 1000
• 1 <= k <= sz

Follow-up: Can you solve the problem in O(1) extra memory space?

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解法1

先找到每次要调换顺序的头尾两个node, 换完顺序之后往后遍历就好了.

/**
 * 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 (k == 1)
            return head;

        ListNode root = new ListNode(-1, head);
        // 偏移量为-1的node 和 起始的node
        ListNode start = root, cur = head;
        while (true) {
            
            // 找到本次计算的最后一个node
            ListNode end = start;
            for (int i = 0; i < k && end != null; i++)
                end = end.next;

            // 找不到符合条件的node, 推出循环
            if (end == null)
                break;

            // 按照顺序依次将node调换顺序
            while (start.next != end) {
                ListNode next = cur.next;
                cur.next = next.next;
                next.next = start.next;
                start.next = next;
            }
            
            // 准备进行下一次循环
            start = cur;
            cur = cur.next;
        }

        return root.next;
    }
}

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解法2

更简洁的写法

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {

        int count = 0;
        ListNode cur = head;
        while (cur != null && ++count < k) {
            cur = cur.next;
        }
        if (count < k)
            return head;

        cur.next = reverseKGroup(cur.next, k);;
        while (head != cur) {
            ListNode temp = head.next;
            head.next = cur.next;
            cur.next = head;

            head = temp;
        }
        return head;
    }
}