LeetCode-328. Odd Even Linked List

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问题描述

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.

Note that the relative order inside both the even and odd groups should remain as it was in the input.

You must solve the problem in O(1) extra space complexity and O(n) time complexity.

Example :

img

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Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

Constraints:

  • n == number of nodes in the linked list
  • 0 <= n <= 104
  • -106 <= Node.val <= 106

解答

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/*
 * @Description: 328. Odd Even Linked List
 * @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
 * @return {ListNode}
 */
const oddEvenList = function (head) {
  if (head === null) {
    return null
  }
  let odd = head
  let even = head.next
  let evenHead = even
  while (even !== null && even.next !== null) {
    odd.next = even.next
    odd = odd.next
    even.next = odd.next
    even = even.next
  }
  odd.next = evenHead
  return head
}