C++编写程序将一有序数组多余的元素删除并输出

Answer 1

#include <stdio.h>
#define ARR_SIZE 5
int Delete(int *a, int n, int x);

void main()
{
int a[ARR_SIZE-1], x, i, n;

printf("Please enter array numbers:\n");
scanf("%d", &n);

printf("Please enter array:\n");
for (i=0; i<n; i++)
{
scanf("%d", &a[i]);
}

printf("Please enter a number what you want to delet:\n");
scanf("%d", &x);

if( Delete(a, n, x) == -1 )
printf("can not find the number!\n") ;
else
{
printf("After delet %d:\n", x);
for(i=0; i < n - 1; i++)
{
printf("%d ", a[i]); // 不要加地址符，我还以为我编译器坏了，囧
}
}
}

int Delete(int *a, int n, int x)
{
int *p=NULL, *pos=NULL;

for (p=a; p<a+n; p++)
{
if (*p == x)
break;
}

if( p >= a + n )
return -1 ;

for( ; p < a + n - 1 ; ++p ) //循环反了，后边的数据会把前边的覆盖
{
*p=*(p+1);
}

return 0 ;
}

Answer 2

#include <iostream>
using namespace std;
int unique(int *a, int n) {
if (n <= 0)
return 0;
int m = 1;
for (int i = 1; i < n; i ++) {
if (a[i] != a[m - 1]) {
a[m ++] = a[i];
}
}
return m;//返回修改后数组的长度
}
int main() {
int n = 11;
int a[] = {1, 1, 2, 3 ,3, 4, 4, 5, 5, 5, 5};
n = unique(a, n);
for (int i = 0; i < n; i ++) {
cout << a[i] << " ";
}
cout << endl;
return 0;
}
或
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int n = 11;
int a[] = {1, 1, 2, 3 ,3, 4, 4, 5, 5, 5, 5};
n = unique(a, a + n) - a;//使用STL自带去重函数，返回值为最后一个元素的位置（地址）
for (int i = 0; i < n; i ++) {
cout << a[i] << " ";
}
cout << endl;
return 0;
}

Answer 3