148. 排序链表（JS实现）

1 题目

在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序。
示例 1:
输入: 4->2->1->3
输出: 1->2->3->4
示例 2:
输入: -1->5->3->4->0
输出: -1->0->3->4->5

2 思路

快速排序的链表方式，注意获取中点时，链表被断开成等分的两半，返回的是各部分的首节点

3代码

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var sortList = function(head) {
  
  if (!head || !head.next) return head;

  return sort(head);

  function sort(linkHead) {
    if (!linkHead || !linkHead.next) return linkHead;
    let [head1, head2] = getMid(linkHead);
    head1 = sort(head1);
    head2 = sort(head2);
    return merge(head1, head2);
  }

  function merge(head1, head2) {
    if (!head1 || !head2) return head1 || head2;

    let linkHead = null;
    let list = null;
    while(head1 && head2) {
      if (head1.val < head2.val) {
        if (!linkHead) {
          linkHead = list = head1;
        } else {
          list.next = head1;
          list = list.next;
        }
        head1 = head1.next;
      } else {
        if (!linkHead) {
          linkHead = list = head2;
        } else {
          list.next = head2;
          list = list.next;
        }
        head2 = head2.next;
      }
    }

    if (head1) {
      list.next = head1;
    }

    if (head2) {
      list.next = head2;
    }

    return linkHead;
  }

  function getMid(linkHead) {
    if (!linkHead) return [null, null];
    let p = q = linkHead;
    let pre;
    while(p && q) {
      pre = p;
      p = p.next;
      q = q.next ? q.next.next : q.next;
    }
    let head1 = pre && pre.next;
    if (pre) pre.next = null;
    return [linkHead, head1];
  }
};