25.reverse-nodes-in-k-groups-cn

题目地址

https://leetcode.com/problems/reverse-nodes-in-k-group/

题目描述

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.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.

思路

题意是以 k 个nodes为一组进行翻转，返回翻转后的linked list.

从左往右扫描一遍linked list，扫描过程中，以k为单位把数组分成若干段，对每一段进行翻转。给定首尾nodes，如何对链表进行翻转。

链表的翻转过程，初始化一个为null的 previous node（prev），然后遍历链表的同时，当前node （curr）的下一个（next）指向前一个node（prev），在改变当前node的指向之前，用一个临时变量记录当前node的下一个node（curr.next). 即

ListNode temp = curr.next;
curr.next = prev;
prev = curr;
curr = temp;

举例如图：翻转整个链表 1->2->3->4->null -> 4->3->2->1->null

这里是对每一组（k个nodes）进行翻转，

先分组，用一个count变量记录当前节点的个数
用一个start 变量记录当前分组的起始节点位置的前一个节点
用一个end变量记录要翻转的最后一个节点位置
翻转一组（k个nodes）即(start, end) - start and end exclusively。
翻转后，start指向翻转后链表, 区间（start，end）中的最后一个节点, 返回start 节点。
如果不需要翻转，end 就往后移动一个（end=end.next)，每一次移动，都要count+1.

如图所示步骤4和5：翻转区间链表区间（start， end）

举例如图，head=[1,2,3,4,5,6,7,8], k = 3

NOTE: 一般情况下对链表的操作，都有可能会引入一个新的dummy node，因为head有可能会改变。这里head 从1->3, dummy (List(0))保持不变。

复杂度分析

时间复杂度: O(n) - n is number of Linked List
空间复杂度: O(1)

关键点分析

创建一个dummy node
对链表以k为单位进行分组，记录每一组的起始和最后节点位置
对每一组进行翻转，更换起始和最后的位置
返回dummy.next.

代码 (`Java/Python3`)

Java Code

class ReverseKGroupsLinkedList {
  public ListNode reverseKGroup(ListNode head, int k) {
      if (head == null || k == 1) {
        return head;
      }
      ListNode dummy = new ListNode(0);
      dummy.next = head;

      ListNode start = dummy;
      ListNode end = head;
      int count = 0;
      while (end != null) {
        count++;
        // group
        if (count % k == 0) {
          // reverse linked list (start, end]
          start = reverse(start, end.next);
          end = start.next;
        } else {
          end = end.next;
        }
      }
      return dummy.next;
    }

    /** 
     * reverse linked list from range (start, end), return last node.
     * for example: 
     * 0->1->2->3->4->5->6->7->8
     * |           |
     * start       end
     * 
     * After call start = reverse(start, end)
     * 
     * 0->3->2->1->4->5->6->7->8
     *          |  |
     *       start end
     *       first
     * @return the reversed list's 'start' node, which is the precedence of node end
     */
    private ListNode reverse(ListNode start, ListNode end) {
      ListNode curr = start.next;
      ListNode prev = start;
      ListNode first = curr;
      while (curr != end){
        ListNode temp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = temp;
      }
      start.next = prev;
      first.next = curr;
      return first;
    }
}

Python3 Cose

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        if head is None or k < 2:
            return head
        dummy = ListNode(0)
        dummy.next = head
        start = dummy
        end = head
        count = 0
        while end:
            count += 1
            if count % k == 0:
                start = self.reverse(start, end.next)
                end = start.next
            else:
                end = end.next
        return dummy.next

    def reverse(self, start, end):
        prev, curr = start, start.next
        first = curr
        while curr != end:
            temp = curr.next
            curr.next = prev
            prev = curr
            curr = temp
        start.next = prev
        first.next = curr
        return first

参考（References)

Leetcode Discussion (yellowstone)

扩展

要求从后往前以k个为一组进行翻转。(字节跳动（ByteDance）面试题)
例子，1->2->3->4->5->6->7->8, k = 3,
从后往前以k=3为一组，
- 6->7->8 为一组翻转为8->7->6，
- 3->4->5为一组翻转为5->4->3.
- 1->2只有2个nodes少于k=3个，不翻转。
  最后返回： 1->2->5->4->3->8->7->6