编写高效的C程序以查找数组中的最小和第二个最小元素。
范例：
Input: arr[] = {12, 13, 1, 10, 34, 1}
Output: The smallest element is 1 and
second Smallest element is 10一个简单的解决方案是按升序对数组进行排序。排序数组中的前两个元素将是两个最小的元素。该解决方案的时间复杂度为O(n Log n)。
更好的解决方案是扫描阵列两次。在第一次遍历中找到最小元素。令此元素为x。在第二遍历中，找到大于x的最小元素。该解决方案的时间复杂度为O(n)。
上面的解决方案需要两次遍历输入数组。
一个有效的解决方案可以在一次遍历中找到最少两个元素。下面是完整的算法。
算法：
1) Initialize both first and second smallest as INT_MAX
first = second = INT_MAX
2) Loop through all the elements.
a) If the current element is smaller than first, then update first
and second.
b) Else if the current element is smaller than second then update
second执行：
C++
// C++ program to find smallest and
// second smallest elements
#include
using namespace std; /* For INT_MAX */
void print2Smallest(int arr[], int arr_size)
{
int i, first, second;
/* There should be atleast two elements */
if (arr_size < 2)
{
cout<<" Invalid Input ";
return;
}
first = second = INT_MAX;
for (i = 0; i < arr_size ; i ++)
{
/* If current element is smaller than first
then update both first and second */
if (arr[i] < first)
{
second = first;
first = arr[i];
}
/* If arr[i] is in between first and second
then update second */
else if (arr[i] < second && arr[i] != first)
second = arr[i];
}
if (second == INT_MAX)
cout << "There is no second smallest element\n";
else
cout << "The smallest element is " << first << " and second "
"Smallest element is " << second << endl;
}
/* Driver code */
int main()
{
int arr[] = {12, 13, 1, 10, 34, 1};
int n = sizeof(arr)/sizeof(arr[0]);
print2Smallest(arr, n);
return 0;
}
// This is code is contributed by rathbhupendra C
// C program to find smallest and second smallest elements
#include
#include /* For INT_MAX */
void print2Smallest(int arr[], int arr_size)
{
int i, first, second;
/* There should be atleast two elements */
if (arr_size < 2)
{
printf(" Invalid Input ");
return;
}
first = second = INT_MAX;
for (i = 0; i < arr_size ; i ++)
{
/* If current element is smaller than first
then update both first and second */
if (arr[i] < first)
{
second = first;
first = arr[i];
}
/* If arr[i] is in between first and second
then update second */
else if (arr[i] < second && arr[i] != first)
second = arr[i];
}
if (second == INT_MAX)
printf("There is no second smallest element\n");
else
printf("The smallest element is %d and second "
"Smallest element is %d\n", first, second);
}
/* Driver program to test above function */
int main()
{
int arr[] = {12, 13, 1, 10, 34, 1};
int n = sizeof(arr)/sizeof(arr[0]);
print2Smallest(arr, n);
return 0;
} Java
// Java program to find smallest and second smallest elements
import java.io.*;
class SecondSmallest
{
/* Function to print first smallest and second smallest
elements */
static void print2Smallest(int arr[])
{
int first, second, arr_size = arr.length;
/* There should be atleast two elements */
if (arr_size < 2)
{
System.out.println(" Invalid Input ");
return;
}
first = second = Integer.MAX_VALUE;
for (int i = 0; i < arr_size ; i ++)
{
/* If current element is smaller than first
then update both first and second */
if (arr[i] < first)
{
second = first;
first = arr[i];
}
/* If arr[i] is in between first and second
then update second */
else if (arr[i] < second && arr[i] != first)
second = arr[i];
}
if (second == Integer.MAX_VALUE)
System.out.println("There is no second" +
"smallest element");
else
System.out.println("The smallest element is " +
first + " and second Smallest" +
" element is " + second);
}
/* Driver program to test above functions */
public static void main (String[] args)
{
int arr[] = {12, 13, 1, 10, 34, 1};
print2Smallest(arr);
}
}
/*This code is contributed by Devesh Agrawal*/Python
# Python program to find smallest and second smallest elements
import sys
def print2Smallest(arr):
# There should be atleast two elements
arr_size = len(arr)
if arr_size < 2:
print "Invalid Input"
return
first = second = sys.maxint
for i in range(0, arr_size):
# If current element is smaller than first then
# update both first and second
if arr[i] < first:
second = first
first = arr[i]
# If arr[i] is in between first and second then
# update second
elif (arr[i] < second and arr[i] != first):
second = arr[i];
if (second == sys.maxint):
print "No second smallest element"
else:
print 'The smallest element is',first,'and' \
' second smallest element is',second
# Driver function to test above function
arr = [12, 13, 1, 10, 34, 1]
print2Smallest(arr)
# This code is contributed by Devesh AgrawalC#
// C# program to find smallest
// and second smallest elements
using System;
class GFG
{
/* Function to print first smallest
and second smallest elements */
static void print2Smallest(int []arr)
{
int first, second, arr_size = arr.Length;
/* There should be atleast two elements */
if (arr_size < 2)
{
Console.Write(" Invalid Input ");
return;
}
first = second = int.MaxValue;
for (int i = 0; i < arr_size ; i ++)
{
/* If current element is smaller than first
then update both first and second */
if (arr[i] < first)
{
second = first;
first = arr[i];
}
/* If arr[i] is in between first and second
then update second */
else if (arr[i] < second && arr[i] != first)
second = arr[i];
}
if (second == int.MaxValue)
Console.Write("There is no second" +
"smallest element");
else
Console.Write("The smallest element is " +
first + " and second Smallest" +
" element is " + second);
}
/* Driver program to test above functions */
public static void Main()
{
int []arr = {12, 13, 1, 10, 34, 1};
print2Smallest(arr);
}
}
// This code is contributed by Sam007PHP
Javascript
输出 ：
The smallest element is 1 and second Smallest element is 10可以使用相同的方法在数组中查找最大和第二大元素。
时间复杂度： O(n)