作者:
晚于: 2020-07-08 12:00:00后提交分数乘系数50%
截止日期: 2020-07-15 12:00:00
问题描述 :
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
可使用以下代码,完成其中的sortList函数,其中形参head指向无头结点单链表。
#include<iostream>
using namespace std;
struct ListNode
{
int val;
ListNode *next;
ListNode() : val(0), next(NULL) {}
ListNode(int x) : val(x), next(NULL) {}
ListNode(int x, ListNode *next) : val(x), next(next) {}
};
class Solution {
public:
ListNode* sortList(ListNode* head)
{
//填充本函数完成功能
}
};
ListNode *createByTail()
{
ListNode *head;
ListNode *p1,*p2;
int n=0,num;
int len;
cin>>len;
head=NULL;
while(n<len && cin>>num)
{
p1=new ListNode(num);
n=n+1;
if(n==1)
head=p1;
else
p2->next=p1;
p2=p1;
}
return head;
}
void displayLink(ListNode *head)
{
ListNode *p;
p=head;
cout<<"head-->";
while(p!= NULL)
{
cout<<p->val<<"-->";
p=p->next;
}
cout<<"tail\n";
}
int main()
{
ListNode* head = createByTail();
head=Solution().sortList(head);
displayLink(head);
return 0;
}
输入说明 :
首先输入链表长度len,然后输入len个整数,以空格分隔。
输出说明 :
输出格式见范例
输入范例 :
5 -1 5 3 4 0
输出范例:
head-->-1-->0-->3-->4-->5-->tail
#include<iostream> using namespace std; struct ListNode { int val; ListNode *next; ListNode() : val(0), next(NULL) {} ListNode(int x) : val(x), next(NULL) {} ListNode(int x, ListNode *next) : val(x), next(next) {} }; class Solution { public: ListNode* sortList(ListNode* head) { ListNode newhead(-1); newhead.next = head; ListNode *p = head; int length = 0; while (p) { ++length; p = p->next; } for (int step = 1; step < length;step<<= 1) { ListNode *cur = newhead.next; ListNode *tail = & newhead; while (cur) { ListNode *left = cur; ListNode *right = cut(left, step); cur = cut(right, step); tail->next = merge(left, right); while (tail->next) { tail = tail->next; } } } return newhead.next; } ListNode* cut(ListNode* head, int n) { ListNode *p = head; while (--n && p) { p = p->next; } if (!p) return NULL; ListNode *next = p->next; p->next = NULL; return next; } ListNode* merge(ListNode* l1, ListNode* l2) { ListNode newhead(-1); ListNode *p = & newhead; while (l1 && l2) { if (l1->val < l2->val) { p->next = l1; p = l1; l1 = l1->next; } else { p->next = l2; p = l2; l2 = l2->next; } } p->next = l1 ? l1 : l2; return newhead.next; } }; ListNode *createByTail() { ListNode *head; ListNode *p1,*p2; int n=0,num; int len; cin>>len; head=NULL; while(n<len && cin>>num) { p1=new ListNode(num); n=n+1; if(n==1) head=p1; else p2->next=p1; p2=p1; } return head; } void displayLink(ListNode *head) { ListNode *p; p=head; cout<<"head-->"; while(p!= NULL) { cout<<p->val<<"-->"; p=p->next; } cout<<"tail\n"; } int main() { ListNode* head = createByTail(); head=Solution().sortList(head); displayLink(head); return 0; }
原文:https://www.cnblogs.com/zmmm/p/13614695.html