友情支持

如果您觉得这个笔记对您有所帮助,看在D瓜哥码这么多字的辛苦上,请友情支持一下,D瓜哥感激不尽,😜

支付宝

微信

有些打赏的朋友希望可以加个好友,欢迎关注D 瓜哥的微信公众号,这样就可以通过公众号的回复直接给我发信息。

wx jikerizhi

公众号的微信号是: jikerizhi因为众所周知的原因,有时图片加载不出来。 如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。

148. 排序链表

给你链表的头结点 head ,请将其按 升序 排列并返回 排序后的链表

示例 1:

0148 01
输入:head = [4,2,1,3]
输出:[1,2,3,4]

示例 2:

0148 02
输入:head = [-1,5,3,4,0]
输出:[-1,0,3,4,5]

示例 3:

输入:head = []
输出:[]

提示:

  • 链表中节点的数目在范围 [0, 5 * 104]

  • -105 <= Node.val <= 105

进阶:你可以在 \(O(n*logn)\) 时间复杂度和常数级空间复杂度下,对链表进行排序吗?

思路分析

  • 一刷

  • 二刷

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
/**
 * Runtime: 4 ms, faster than 62.99% of Java online submissions for Sort List.
 *
 * Memory Usage: 40.2 MB, less than 89.47% of Java online submissions for Sort List.
 *
 * @author D瓜哥 · https://www.diguage.com
 * @since 2019-10-29 21:05
 */
public ListNode sortList(ListNode head) {
    if (Objects.isNull(head) || Objects.isNull(head.next)) {
        return head;
    }
    ListNode prev = null;
    ListNode slow = head;
    ListNode fast = head;
    while (Objects.nonNull(fast) && Objects.nonNull(fast.next)) {
        prev = slow;
        slow = slow.next;
        fast = fast.next.next;
    }
    prev.next = null;

    ListNode firstResult = sortList(head);
    ListNode secondResult = sortList(slow);

    return merge(firstResult, secondResult);
}

private ListNode merge(ListNode first, ListNode second) {
    ListNode result = new ListNode(0);
    ListNode tail = result;
    while (Objects.nonNull(first) && Objects.nonNull(second)) {
        if (first.val < second.val) {
            tail.next = first;
            first = first.next;
        } else {
            tail.next = second;
            second = second.next;
        }
        tail = tail.next;
    }
    if (Objects.isNull(first)) {
        tail.next = second;
    }
    if (Objects.isNull(second)) {
        tail.next = first;
    }
    return result.next;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
/**
 * @author D瓜哥 · https://www.diguage.com
 * @since 2025-04-08 15:06:31
 */
public ListNode sortList(ListNode head) {
  if (head == null || head.next == null) {
    return head;
  }
  // 分
  ListNode pre = head, slow = head, fast = head;
  while (fast != null && fast.next != null) {
    pre = slow;
    slow = slow.next;
    fast = fast.next.next;
  }
  pre.next = null; // 将链表切成两段
  // 治
  ListNode r1 = sortList(head);
  ListNode r2 = sortList(slow);
  // 合
  return merge(r1, r2);
}

private ListNode merge(ListNode l1, ListNode l2) {
  ListNode result = new ListNode(0);
  ListNode cur = result;
  while (l1 != null && l2 != null) {
    if (l1.val <= l2.val) {
      cur.next = l1;
      l1 = l1.next;
    } else {
      cur.next = l2;
      l2 = l2.next;
    }
    cur = cur.next;
  }
  if (l1 != null) {
    cur.next = l1;
  }
  if (l2 != null) {
    cur.next = l2;
  }
  return result.next;
}

参考资料