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  1. K 个一组翻转链表

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]

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

提示:
链表中的节点数目为 n
1 <= k <= n <= 5000
0 <= Node.val <= 1000

进阶:你可以设计一个只用 O(1) 额外内存空间的算法解决此问题吗?

javascript
/**
 * 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}
 */
var reverseKGroup = function(head, k) {
  let a = b = head
  for (let i = 0; i < k; i++) {
    if(b === null) return head
    b = b.next
  }
  const newHead = reverse(a, b)
  a.next = reverseKGroup(b, k)

  return newHead
}

function reverse(a, b){
  let pre = null, cur = a, next = a
  while(cur !== b){
    next = cur.next
    cur.next = pre
    pre = cur
    cur = next
  }
  return pre
}

// test
function ListNode(val, next) {
  this.val = val === undefined ? 0 : val
  this.next = next === undefined ? null : next
}

var l1 = new ListNode(1)
var lstNode1 = new ListNode(2)
var lstNode2 = new ListNode(3)
var lstNode3 = new ListNode(4)
l1.next = lstNode1
lstNode1.next = lstNode2
lstNode2.next = lstNode3

console.log(reverseKGroup(l1, 3))