麻煩了，發現鏈表的排序是個挺麻煩的問題。
剛才搞了一個多鐘都出不來。
剛才搞了一個多鐘都出不來。
#include<iostream>
#include<string>
using namespace std;
typedef struct phone
{
string data;
int times;
struct phone *next;
}node;
char change(char one)
{
if(one>='A'&&one<='Z')
{
switch(one)
{
case 'A':
case 'B':
case 'C':
return '2';
break;
case 'D':
case 'E':
case 'F':
return '3';
break;
case 'G':
case 'H':
case 'I':
return '4';
break;
case 'J':
case 'K':
case 'L':
return '5';
break;
case 'M':
case 'N':
case 'O':
return '6';
break;
case 'P':
case 'R':
case 'S':
return '7';
break;
case 'T':
case 'U':
case 'V':
return '8';
break;
case 'W':
case 'X':
case 'Y':
return '9';
break;
}
}
else return one;
}
main()
{
node *head,*pre,*p,*q,*l,*Qpre,*Ppre;
int x,y,z,i,k;
string temp1,temp2;
cout<<"enter a number"<<endl;
cin >>x;
head=new node;
pre=head;
for(i=0;i<x;i++)
{
cin >>temp1;
y=temp1.length();
k=0;
temp2="";
for(z=0;z<y;z++)
{
if(k==3)
{
temp2=temp2+'-';
k++;
}
if(temp1[z]!='-')
{
temp2=temp2+change(temp1[z]);
k++;
}
}
p=new node;
p->data=temp2;
pre->next=p;
pre=pre->next;
}
pre->next=NULL;
cout<<endl<<endl;
pre=head->next;
l=head;
while(pre)
{
temp1=pre->data;
p=pre->next;
q=pre;
k=1;
while(p)
{
if(temp1==p->data)
{
k++;
q->next=p->next;
delete(p);
p=q->next;
}
else
{
p=p->next;
q=q->next;
}
}
if(k>1)
{
pre->times=k;
l=pre;
pre=pre->next;
}
else
{
l->next=pre->next;
delete(pre);
pre=l->next;
}
}
pre=head;
while(pre)
{
p=pre->next;
q=p;
Qpre=Ppre=pre;
while(p)
{
if(p->data<q->data)
{
Qpre=Ppre;
q=p;
p=p->next;
Ppre=Ppre->next;
}
else
{
Ppre=Ppre->next;
p=p->next;
}
}
cout<<q->data<<' '<<q->times<<endl;
Qpre->next=q->next;
delete(q);
pre=pre->next;
}
cout<<head->next->data<<' '<<head->next->times<<endl;
return 0;
}
終于出來了啦。樓主你試試。
