H面试程序(16): 简单选择排序

 

#include<stdio.h>
#include<assert.h>

void display(int * a, int n) 
{   
	assert(a);
	for(int i = 0; i < n; i++)
	{
		printf("%d,",a[i]);
	}
	printf("\n");
}

void swap(int * a, int * b)
{   
	assert(a);
	assert(b);
	int temp;
	temp = *a;
	*a = *b;
	*b = temp;
}

void select_sort(int * a, int n)
{  
	assert(a);
	int i, j,min;
	for( i = 0; i < n-1  ; i++) //i 0 ~ n-2
	{
		min = a[i]; //每次选择一个作为最小值
	
		for( j = i; j < n; j++) // j i ~ n-1
		{
			if(a[j] < min)
			{
			   swap(&(a[j]), &(a[i]));  
			}
		}
	}
}

int main()
{   
	int a[10] ={2, 1, 3, 4, 5, 7,2,3, 6, 1};
	int num = sizeof(a)/sizeof(int);
	printf("before_sort:");
	display(a, num);
	select_sort(a,num);
	printf("after_sort:");
	display(a, num);
	return 0;
	
}


 

#include <stdio.h>
#include <time.h>

//交换两个数据
void swap(int *a, int *b)
{
	int temp;
	temp = *a;
	*a = *b;
	*b = temp;
}


//显示交换后的数组
void display_array( int a[], int n )
{
    int i;
    for( i = 0; i < n; i++ )
        printf( "%d ", a[i] );
}



//选择排序
void select_sort( int a[], int n )
{

	int i ,j;
	
	for(i = 0; i<n-1 ; i++ )    
	{                                  
	    int min = a[i];     //每次都设定最小的数,i 0 ~ 8
		for(j =i+1; j <n; j++)  // 将a[i]~ a[9]的数于当前最小的数进行比较,如果小于当前最小数就交换
		{
			if(a[j] < min)
			{	
				swap(&(a[j]), &(a[i]));
			}
		}
	}

}


int main()
{
	clock_t start, finish;
	start = clock();

    int  n = 10;
    int a[] = { 2, 1, 3, 4, 5, 7, 6, 8 ,2,4};
    printf( "Before sorting: " );
    display_array( a, n );
    

    select_sort( a, n );
    printf( "After sorting: " );
    display_array( a, n );
    finish = clock();
    printf("\n本次计算一共耗时: %f秒\n\n", (double)(finish-start)/CLOCKS_PER_SEC);
    return 0;
}


 

 

posted @ 2013-09-15 18:42  pangbangb  阅读(151)  评论(0编辑  收藏  举报