面试算法-164-K 个一组翻转链表

题目

给你链表的头节点 head ，每 k 个节点一组进行翻转，请你返回修改后的链表。

k 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。

你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。

示例 1：

输入：head = [1,2,3,4,5], k = 2
输出：[2,1,4,3,5]

解

/**
 * 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 dummy = new ListNode(0);
        dummy.next = head;
        int count = 0;
        ListNode pre = dummy;
        ListNode cur = head;
        ListNode next = null;
        while (cur != null) {
            next = cur.next;
            count++;

            if (count == k) {
                ListNode p1 = pre.next;
                ListNode p2 = cur;
                p2.next = null;

                reverse(p1);

                pre.next = p2;
                p1.next = next;

                pre = p1;
                count = 0;
            }
            cur = next;
        }
        return dummy.next;
    }

    public void reverse(ListNode head) {
        if (head == null || head.next == null) {
            return;
        }

        ListNode pre = null;
        ListNode cur = head;
        ListNode next = null;
        while (cur != null) {
            next = cur.next;

            cur.next = pre;

            pre = cur;
            cur = next;
        }
    }
}