数据结构实验之链表七:单链表中重复元素的删除
Time Limit: 1000MS Memory Limit: 65536KB
Problem Description
按照数据输入的相反顺序(逆位序)建立一个单链表,并将单链表中重复的元素删除(值相同的元素只保留最后输入的一个)。
Input
第一行输入元素个数 n (1 <= n <= 15);
第二行输入 n 个整数,保证在 int 范围内。
Output
第一行输出初始链表元素个数;
第二行输出按照逆位序所建立的初始链表;
第三行输出删除重复元素后的单链表元素个数;
第四行输出删除重复元素后的单链表。
Example Input
10 21 30 14 55 32 63 11 30 55 30
Example Output
10 30 55 30 11 63 32 55 14 30 21 7 30 55 11 63 32 14 21
Hint
Author
不得使用数组!
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct node
{
int data;
struct node * next;
};
int main()
{
int n,i;
struct node *p, *q, *head, *tail;
head = (struct node *)malloc(sizeof(struct node));
head->next = NULL;
scanf("%d", &n);
for(i = 0; i < n; i++)
{
p = (struct node *)malloc(sizeof(struct node));
scanf("%d", &p->data);
p->next = head->next;
head->next = p;
}
q = head->next;
printf("%d\n", n);
for(i = 0; i < n; i++)
{
printf("%d", q->data);
if(i==n-1)
printf("\n");
else
printf(" ");
q = q->next;
}
for(q = head->next; q->next!=NULL;)
{
for(tail = q,p=tail->next; p!=NULL; )
{
if(p->data==q->data)
{
tail->next = p->next;
n--;
break;
}
else
{
tail = tail->next;
p = tail->next;
}
}
if(p!=NULL)
{
q = q;
}
else
q = q->next;
}
printf("%d\n", n);
for(q = head->next; q != NULL; q=q->next)
{
printf("%d", q->data);
if(q->next == NULL)
{
printf("\n");
}
else
printf(" ");
}
return 0;
}
再上个c++的 #include<bits/stdc++.h>
using namespace std;
struct node
{
int data;
struct node *next;
};
int main()
{
int n, i;
struct node *head, *t, *p, *q;
cin>>n;
head = new node;
head->next = NULL;
q = head;
for(i = 0; i < n; i++)
{
p = new node;
cin>>p->data;
p->next = head->next;
head->next = p;
}
q = head->next;
cout<<n<<endl;
while(q)
{
cout<<q->data;
if(q->next)
cout<<" ";
else
cout<<endl;
q = q->next;
}
for(q = head->next;q->next;)
{
for(t = q,p = t->next;p;)
{
if(q->data == p->data)
{
t->next = p->next;
n--;
break;
}
else
{
t = t->next;
p = p->next;
}
}
if(p)
q = q;
else
q = q->next;
}
cout<<n<<endl;
q = head->next;
while(q)
{
cout<<q->data;
if(q->next)
cout<<" ";
else
cout<<endl;
q = q->next;
}
return 0;
}