ReverseNodesInKGroup.java

/**
 * <p>给你一个链表，每 <em>k </em>个节点一组进行翻转，请你返回翻转后的链表。</p>
 *
 * <p><em>k </em>是一个正整数，它的值小于或等于链表的长度。</p>
 *
 * <p>如果节点总数不是 <em>k </em>的整数倍，那么请将最后剩余的节点保持原有顺序。</p>
 *
 * <p><strong>进阶：</strong></p>
 *
 * <ul>
 * <li>你可以设计一个只使用常数额外空间的算法来解决此问题吗？</li>
 * <li><strong>你不能只是单纯的改变节点内部的值</strong>，而是需要实际进行节点交换。</li>
 * </ul>
 *
 * <p> </p>
 *
 * <p><strong>示例 1：</strong></p>
 * <img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg" style="width: 542px; height: 222px;" />
 * <pre>
 * <strong>输入：</strong>head = [1,2,3,4,5], k = 2
 * <strong>输出：</strong>[2,1,4,3,5]
 * </pre>
 *
 * <p><strong>示例 2：</strong></p>
 * <img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg" style="width: 542px; height: 222px;" />
 * <pre>
 * <strong>输入：</strong>head = [1,2,3,4,5], k = 3
 * <strong>输出：</strong>[3,2,1,4,5]
 * </pre>
 *
 * <p><strong>示例 3：</strong></p>
 *
 * <pre>
 * <strong>输入：</strong>head = [1,2,3,4,5], k = 1
 * <strong>输出：</strong>[1,2,3,4,5]
 * </pre>
 *
 * <p><strong>示例 4：</strong></p>
 *
 * <pre>
 * <strong>输入：</strong>head = [1], k = 1
 * <strong>输出：</strong>[1]
 * </pre>
 *
 * <ul>
 * </ul>
 *
 * <p><strong>提示：</strong></p>
 *
 * <ul>
 * <li>列表中节点的数量在范围 <code>sz</code> 内</li>
 * <li><code>1 <= sz <= 5000</code></li>
 * <li><code>0 <= Node.val <= 1000</code></li>
 * <li><code>1 <= k <= sz</code></li>
 * </ul>
 * <div><div>Related Topics</div><div><li>递归</li><li>链表</li></div></div><br><div><li>👍 1443</li><li>👎 0</li></div>
 */

package leetcode1;

public class ReverseNodesInKGroup {

    public static void main(String[] args) {
        Solution solution = new ReverseNodesInKGroup().new Solution();
    }

    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;
        }
    }

    /**
     * 参考题解，画图看一看就懂了，挺简单
     * https://leetcode-cn.com/problems/reverse-nodes-in-k-group/solution/tu-jie-kge-yi-zu-fan-zhuan-lian-biao-by-user7208t/
     */
    class Solution {
        public ListNode reverseKGroup(ListNode head, int k) {
            ListNode prev = new ListNode();
            prev.next = head;
            head = prev;
            while (true) {
                ListNode end = prev;
                for (int i = 0; i < k && end != null; i++) end = end.next;
                if (end == null) return head.next;
                ListNode start = prev.next;
                ListNode next = end.next;
                end.next = null;
                prev.next = reverse(start);
                start.next = next;
                prev = start;
            }
        }

        private ListNode reverse(ListNode node) {
            ListNode prev = null;
            while (node != null) {
                ListNode next = node.next;
                node.next = prev;
                prev = node;
                node = next;
            }
            return prev;
        }
    }
}