LeetCode-25. Reverse Nodes in K-Group

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问题描述

Given the head of a linked list, reverse the nodes of the list k at a time, and return the 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:

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

Constraints:

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

解答

方法一：递归
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/*
 * @Description: 链表中的节点每k个一组翻转
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
const reverseKGroup = function (head, k) {
  let tail = head
  for (let i = 0; i < k; i++) {
    if (tail === null) {
      return head
    }
    tail = tail.next
  }

  let pre = null
  let cur = head
  while (cur !== tail) {
    let temp = cur.next
    cur.next = pre
    pre = cur
    cur = temp
  }
  head.next = reverseKGroup(tail, k)
  return pre
}
方法二：迭代反转每一组子链表
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/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
let reverseKGroup2 = function (head, k) {
  let length = getLength(head)
  let newHead = head
  for (let i = 1; (i + k - 1) <= length; i += k) {
    newHead = reverseBetween(newHead, i, i + k - 1)
  }
  return newHead
}

function getLength (head) {
  let n = 0
  let cur = head
  while (cur !== null) {
    cur = cur.next
    n++
  }
  return n
}

function reverseBetween (head, left, right) {
  let dummy = new ListNode(-1)
  dummy.next = head
  let pre = dummy
  let cur = head
  for (let i = 1; i < left; i++) {
    pre = cur
    cur = cur.next
  }
  for (let j = left; j < right; j++) {
    let temp = cur.next
    cur.next = temp.next
    temp.next = pre.next
    pre.next = temp
  }
  return dummy.next
}