当前位置：首页 > news >正文

开网站做什么/优化服务平台

news 2025/7/8 3:31:42

开网站做什么,优化服务平台,免费前端模板,徐州最新消息今天在 O(n log n) 时间复杂度和常数级的空间复杂度下给链表排序。您在真实的面试中是否遇到过这个题？ Yes样例给出 1->3->2->null，给它排序变成 1->2->3->null. 这道题目明显是需要我们用归并进行链表排序。 C Code 12345678910111213…

在 O(n log n) 时间复杂度和常数级的空间复杂度下给链表排序。

您在真实的面试中是否遇到过这个题？

Yes

样例

给出 1->3->2->null，给它排序变成 1->2->3->null.

这道题目明显是需要我们用归并进行链表排序。

C++ Code 

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
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70

/**
* Definition of ListNode
* class ListNode {
* public:
*     int val;
*     ListNode *next;
*     ListNode(int val) {
*         this->val = val;
*         this->next = NULL;
*     }
* }
*/
class Solution
{
public:
    /**
     * @param head: The first node of linked list.
     * @return: You should return the head of the sorted linked list,
                    using constant space complexity.
     */
    ListNode *sortList(ListNode *head)
    {
        // write your code here
        if(head == NULL || head -> next == NULL) return head;
        ListNode *mid = findMid(head);
        ListNode *temp = mid -> next;//将链表分成两段
        mid -> next = NULL;
        ListNode *left = sortList(head);
        ListNode *right = sortList(temp);
        return merge(left, right);
    }
    ListNode *findMid(ListNode *head)//寻找链表中间结点
    {
        if(head == NULL || head -> next == NULL) return head;
        ListNode *fast = head;
        ListNode *slow = head;
        while(fast -> next != NULL && fast -> next -> next != NULL)
        {
            fast = fast -> next -> next;
            slow = slow -> next;
        }
        return slow;
    }
    ListNode *merge(ListNode *left, ListNode *right)//合并链表
    {
        if(left == NULL && right == NULL) return NULL;
        ListNode *node = new ListNode(0);
        ListNode *node1 = node;
        while(left != NULL && right != NULL)
        {
            if(left -> val < right -> val)
            {
                ListNode *temp = left;
                left = left -> next;
                node1 -> next = temp;
                node1 = node1 -> next;
            }
            else
            {
                ListNode *temp = right;
                right = right -> next;
                node1 -> next = temp;
                node1 = node1 -> next;
            }
        }
        if(left != NULL) node1 -> next = left;
        else node1 -> next = right;
        return node -> next;
    }
};

查看全文

http://www.jmfq.cn/news/4757545.html