【一天一大 lee】重排链表 (难度:中等) - Day20201020

给定链表 1->2->3->4, 重新排列为 1->4->2->3.

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

/**
 * 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) {
  if (head == null) return
  // 翻转链表
  function reverseList(root) {
    let _result = null,
      node = root
    while (node != null) {
      // 逐个节点与其next指针上的节点nextTemp交换 -> 翻转链表
      let nextTemp = node.next
      node.next = _result
      _result = node
      node = nextTemp
    }
    return _result
  }

  let fast = head,
    beforeList = head,
    afterList = head
  // 将afterList指针指向后一半节点
  while (fast.next != null && fast.next.next != null) {
    afterList = afterList.next
    fast = fast.next.next
  }
  // 注意翻转后一半链表时会将原链表与后一半链表衔接的next指针置为null切断链表
  afterList = reverseList(afterList)

  let before = null,
    after = null
  while (beforeList != null && afterList != null) {
    before = beforeList.next
    after = afterList.next

    beforeList.next = afterList
    beforeList = before

    afterList.next = beforeList
    afterList = after
  }

  return head
}

var reorderList = function(head) {
  if (head == null) return
  // 存储链表节点
  let list = []
  while (head !== null) {
    let temp = head.next
    head.next = null
    list.push(head)
    head = temp
  }
  // 按顺序重构next指针
  let start = 0,
    end = list.length - 1
  while (start < end) {
    list[start].next = list[end]
    // 数组前后两个指针不相邻时构建next
    if (end - start != 1) list[end].next = list[start + 1]
    start++
    end--
  }
  return list[0]
}