要求算法复杂度不能是O(n^2)
谢谢!
可以先用快速排序进行排序,其中用另外一个进行地址查找
代码如下,在VC++6.0运行通过。给分吧^-^
//快速排序
#include
usingnamespacestd;
intPartition (int*L,intlow,int high)
{
inttemp = L[low];
intpt = L[low];
while (low < high)
{
while (low < high && L[high] >= pt)
--high;
L[low] = L[high];
while (low < high && L[low] <= pt)
++low;
L[low] = temp;
}
L[low] = temp;
returnlow;
}
voidQSort (int*L,intlow,int high)
{
if (low < high)
{
intpl = Partition (L,low,high);
QSort (L,low,pl - 1);
QSort (L,pl + 1,high);
}
}
intmain ()
{
intnarry[100],addr[100];
intsum = 1,t;
cout << "Input number:" << endl;
cin >> t;
while (t != -1)
{
narry[sum] = t;
addr[sum - 1] = t;
sum++;
cin >> t;
}
sum -= 1;
QSort (narry,1,sum);
for (int i = 1; i <= sum;i++)
cout << narry[i] << '/t';
cout << endl;
intk;
cout << "Please input place you want:" << endl;
cin >> k;
intaa = 1;
intkk = 0;
for (;;)
{
if (aa == k)
break;
if (narry[kk] != narry[kk + 1])
{
aa += 1;
kk++;
}
}
cout << "The NO." << k << "number is:" << narry[sum - kk] << endl;
cout << "And it's place is:" ;
for (i = 0;i < sum;i++)
{
if (addr[i] == narry[sum - kk])
cout << i << '/t';
}
return0;
}
1、找错
Void test1()
{
char string[10];
char* str1="0123456789";
strcpy(string, str1);// 溢出,应该包括一个存放'/0'的字符string[11]
}
Void test2()
{
char string[10], str1[10];
for(I=0; I<10;I++)
{
str1[i] ='a';
}
strcpy(string, str1);// I,i没有声明。
}
Void test3(char* str1)
{
char string[10];
if(strlen(str1)<=10)// 改成<10,字符溢出,将strlen改为sizeof也可以
{
strcpy(string, str1);
}
}
2.
void g(int**);
int main()
{
int line[10],i;
int *p=line; //p是地址的地址
for (i=0;i<10;i++)
{
*p=i;
g(&p);//数组对应的值加1
}
for(i=0;i<10;i++)
printf("%d/n",line[i]);
return 0;
}
void g(int**p)
{
(**p)++;
(*p)++;// 无效
}
输出:
1
2
3
4
5
6
7
8
9
10
3. 写出程序运行结果
int sum(int a)
{
auto int c=0;
static int b=3;
c+=1;
b+=2;
return(a+b+c);
}
void main()
{
int I;
int a=2;
for(I=0;I<5;I++)
{
printf("%d,", sum(a));
}
}
// static会保存上次结果,记住这一点,剩下的自己写
输出:8,10,12,14,16,
4.
int func(int a)
{
int b;
switch(a)
{
case 1: 30;
case 2: 20;
case 3: 16;
default: 0
}
return b;
}
则func(1)=?
// b定义后就没有赋值。
5:
int a[3];
a[0]=0; a[1]=1; a[2]=2;
int *p, *q;
p=a;
q=&a[2];
则a[q-p]=a[2]
解释:指针一次移动一个int但计数为1