链表排序 -lintcode
来源:互联网 发布:tnt网络瘫痪 编辑:程序博客网 时间:2024/05/21 09:06
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。
样例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
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
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;
}
};
* 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;
}
};
阅读全文
0 0
- LintCode : 链表排序
- lintcode,链表排序
- LintCode 链表排序
- 链表排序-LintCode
- lintcode--链表排序
- 链表排序 -lintcode
- lintcode链表排序
- LintCode 链表插入排序
- lintcode-链表插入排序
- LintCode-链表插入排序
- 链表插入排序 lintcode
- LintCode 98-链表排序
- LintCode 链表插入排序
- 链表插入排序-LintCode
- LintCode 链表插入排序
- [LintCode]98.链表排序
- 链表插入排序-LintCode
- LintCode 合并两个排序链表
- Android 调用系统邮箱
- windows和linux之间通过scp进行文件复制
- c 语言小技巧之二 自动创建目录
- awk之原文编辑(类似 sed -i 功能)
- 成为Java GC专家(5)—Java性能调优原则
- 链表排序 -lintcode
- JavaScript Tween算法及缓动效果
- Linux常用命令
- HDU1248 寒冰王座(母函数)
- javascript内置对象常用属性和方法(笔记二)
- Android ANR
- spring boot maven继承parent的两种方法
- IAT和导入表
- 使用云加速 Yocto 编译