按原始顺序打印给定数组中的n个最小元素
假设有k个元素数组，程序必须按它们出现的顺序在其中找到n个最小的元素。
Input : arr[] = {1, 2, 4, 3, 6, 7, 8}, k=3
Ouput : 1, 2, 3
Input k is 3 it means 3 shortest elements among the set needs to be displayed in original order like 1 than 2 and than 3算法
START
Step 1 -> start variables as int i, max, pos, j, k=4 and size for array size
Step 2 -> Loop For i=k and i<size and i++
Set max = arr[k-1]
pos = k-1
Loop For j=k-2 and j>=0 and j--
If arr[j]>max
Set max = arr[j]
Set pos = j
End
End
IF max> arr[i]
Set j = pos
Loop While j < k-1
Set arr[j] = arr[j+1]
Set j++
End
Set arr[k-1] = arr[i]
End IF
End
Step 3 -> Loop For i = 0 and i < k and i++
Print arr[i]
STOP示例
#include <stdio.h>
int main() {
int arr[] = {5,8,3,1,2,9};
int i, max, pos, j, k=4;
int size = sizeof(arr)/sizeof(arr[0]);
//使用插入排序，从k开始。
for(i=k;i<size;i++){
max = arr[k-1];
pos = k-1;
for(j=k-2;j>=0;j--) {
if(arr[j]>max) {
max = arr[j];
pos = j;
}
}
if ( max> arr[i] ) {
j = pos;
while( j < k-1 ) {
arr[j] = arr[j+1];
j++;
}
arr[k-1] = arr[i];
}
}
//打印前k个元素
for (i = 0; i < k; i++) {
printf("%d ", arr[i]);
}
return 0;
}输出结果
如果我们运行上面的程序，那么它将生成以下输出。
5 3 1 2