题解 | #链表中的节点每k个一组翻转#

这一题的最大的难点应该就是考虑边界情况了

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    private void reverse(ListNode start, ListNode end) {
        ListNode pre = end.next;
        ListNode c = start;
        ListNode next = start.next;
        
        while(pre != end) {
            c.next = pre;
            pre = c;
            c = next;
            if (next != null) {
                next = next.next;
            }
        }
    }
    
    public ListNode reverseKGroup (ListNode head, int k) {
        // write code here
        if (k == 1) {
            return head;
        }
        ListNode result = head, tail = head;
        ListNode end = head;
        int c = 1;
        while (end != null && end.next != null) {
            end = end.next;
            c++;
            if (c % k == 0) {
                reverse(head, end);
                if (c == k) {
                    // 首次逆序拿到头节点
                    result = end;
                } else {
                    // 尾节点接上本次的逆序头节点
                    tail.next = end;
                }
                // 更新尾节点
                tail = head;
                // 重新开始一个K组
                head = head.next;
                end = head;
                c++;
            }
        }
        return result;
    }
}