Odd Even Linked List

Link

https://leetcode.com/problems/odd-even-linked-list/

Deadline

Dec 13, 2022

Status

Archived

Type

linked list

문제

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.

Constraints:

The number of nodes in the linked list is in the range [0, 104].

106 <= Node.val <= 106

풀이

효성


/**
 * 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}
 */
var oddEvenList = function (head) {
  if (!head) return head;

  let odd = head;
  let even = head.next;
  
  while (odd.next && odd.next.next) {
    const tmp = odd.next;
    odd.next = odd.next.next;
    odd = odd.next;
    tmp.next = odd.next;
  }
  odd.next = even;

  return head;
}

재영


/**
 * 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}
 */

function oddEvenList(head) {
  let copyhead = head;

  const odds = [];
  const evens = [];

  for (let [odd, even] of genOddEvenPairs()) {
    odds.push(odd);
    evens.push(even);
  }
  
  odds[0] = head;
  reorderOddEvenList();

  function* genOddEvenPairs() {
    while (copyhead !== null) {
        yield [copyhead, copyhead.next];
        
        if (!copyhead.next) break;
        copyhead = copyhead.next.next;
    }
  }

  function reorderOddEvenList() {
    odds.concat(evens).map(updateNodeNextNull).forEach((node, idx, origin) => {
        if (idx !== origin.length - 1) {
            node.next = origin[idx + 1]
        }
    })
  }

  function updateNodeNextNull(v) {
    if (v?.next) {
        v.next = null;
    }
    
    return v;
  }

  return head
}