leetcode刷题记录(13)-中等

1.重排链表

题目：

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

思路：先放进数组里，用数组下标去获取节点

/**
 * 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 {void} Do not return anything, modify head in-place instead.
 */
var reorderList = function(head) {
      const list = [];
  while (head) {
    list.push(head);
    head = head.next;
  }
  const pre = new ListNode();
  let node = pre;
  let count = 0;
  while (list.length) {
    if (count % 2) {
      node.next = list.pop();
    } else {
      node.next = list.shift();
    }
    node = node.next;
    count++;
  }
  node.next=null
  return pre.next;
};

可以考虑优化。看新链表的节点顺序，可以从中间截断，分成两个链表，然后后半部分的链表倒置，最后左右链表依次取第一个节点拼接

/**
 * 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 {void} Do not return anything, modify head in-place instead.
 */
var reorderList = function(head) {
  const dummy = new ListNode(0)
  dummy.next = head

  let slow = dummy
  let quick = dummy

  while (quick && quick.next) {
    slow = slow.next
    quick = quick.next
    quick = quick.next
  }

  let right = slow.next
  slow.next = null
  let left = dummy.next

  right = reverseList(right)

  while (left && right) {
    let lNext = lef