leetcode：奇偶链表

描述：

题解：

这个问题最直接的想法就是双指针思想来构建奇偶链表，然后把偶链表连接在奇链表的尾部。

ES6

const oddEvenList = (head) => {
  if (head == null) {
    return null
  }
  let odd = head, even = head.next, 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
};

时间复杂度：O(n)