题解 | #有序序列插入一个数#
有序序列插入一个数
https://www.nowcoder.com/practice/74486aec6fe14d44b509efabf265ee66
方法一：通过冒泡排序的思想
#include <stdio.h>
void add_num(int* arr, int n, int m)
{
arr[n] = m;
int i = 0;
int j = 0;
for (i = 0; i <= n-1; i++)
{
for (j = 0; j <= n - i - 1; j++)
{
if (arr[j] > arr[j + 1])
{
int tmp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = tmp;
}
}
}
for (i = 0; i <= n; i++)
{
printf("%d ",arr[i]);
}
}
int main()
{
int n = 0;
int arr[50] = { 0 };
int i = 0;
scanf("%d",&n);
for (i = 0; i < n; i++)
{
scanf("%d",&arr[i]);
}
int m = 0;
scanf("%d",&m);
add_num(arr, n, m);
return 0;
}
方法二：
#include <stdio.h>
int main()
{
int arr[20] = { 0 };
int n = 0;
int i = 0;
int tmp = 0;
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%d", &arr[i]);
}
int m = 0;
scanf("%d", &m);
for (i = n-1; i >=0; i--)
{
if (m < arr[i])
{
arr[i + 1] = arr[i];
arr[i] = m;
//arr[i] = arr[i + 1];
}
else
break;
}
arr[i + 1] = m;
for (int i = 0;i <= n;i++)
{
printf("%d ", arr[i]);
}
}
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