在数组中找到第一个，第二个和第三个最小元素 · GeeksForGeeks 中文系列教程

# 在数组中找到第一个，第二个和第三个最小元素 > 原文： [https://www.geeksforgeeks.org/find-the-first-second-and-third-minimum-elements-in-an-array/](https://www.geeksforgeeks.org/find-the-first-second-and-third-minimum-elements-in-an-array/) 在`O(n)`中找到数组中的第一，第二和第三最小元素。例子： ``` Input : 9 4 12 6 Output : First min = 4 Second min = 6 Third min = 9 Input : 4 9 1 32 12 Output : First min = 1 Second min = 4 Third min = 9 ``` **第一种方法**：首先，我们可以使用常规方法对数组进行排序，然后打印数组的第一，第二和第三元素。该解决方案的时间复杂度为`O(N log N)`。 **第二种方法**：此解决方案的时间复杂度为`O(n)`。算法- ``` First take an element then if array[index] < Firstelement Thirdelement = Secondelement Secondelement = Firstelement Firstelement = array[index] else if array[index] < Secondelement Thirdelement = Secondelement Secondelement = array[index] else if array[index] < Thirdelement Thirdelement = array[index] then print all the element ``` ## C++ ```cpp // CPP program to find the first, second // and third minimum element in an array #include<bits/stdc++.h> #define MAX 100000 using namespace std; int Print3Smallest(int array[], int n) { int firstmin = MAX, secmin = MAX, thirdmin = MAX; for (int i = 0; i < n; i++) { /* Check if current element is less than firstmin, then update first, second and third */ if (array[i] < firstmin) { thirdmin = secmin; secmin = firstmin; firstmin = array[i]; } /* Check if current element is less than secmin then update second and third */ else if (array[i] < secmin) { thirdmin = secmin; secmin = array[i]; } /* Check if current element is less than then update third */ else if (array[i] < thirdmin) thirdmin = array[i]; } cout << "First min = " << firstmin << "\n"; cout << "Second min = " << secmin << "\n"; cout << "Third min = " << thirdmin << "\n"; } // Driver code int main() { int array[] = {4, 9, 1, 32, 12}; int n = sizeof(array) / sizeof(array[0]); Print3Smallest(array, n); return 0; } ``` ## Java ```java // Java program to find the first, second // and third minimum element in an array import java.util.*; public class GFG { static void Print3Smallest(int array[], int n) { int firstmin = Integer.MAX_VALUE; int secmin = Integer.MAX_VALUE; int thirdmin = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { /* Check if current element is less than firstmin, then update first, second and third */ if (array[i] < firstmin) { thirdmin = secmin; secmin = firstmin; firstmin = array[i]; } /* Check if current element is less than secmin then update second and third */ else if (array[i] < secmin) { thirdmin = secmin; secmin = array[i]; } /* Check if current element is less than then update third */ else if (array[i] < thirdmin) thirdmin = array[i]; } System.out.println("First min = " + firstmin ); System.out.println("Second min = " + secmin ); System.out.println("Third min = " + thirdmin ); } // Driver code public static void main(String[] args) { int array[] = {4, 9, 1, 32, 12}; int n = array.length; Print3Smallest(array, n); } } // This code is contributed by // Sam007 ``` ## Python3 ```py # A Python program to find the first, # second and third minimum element # in an array MAX = 100000 def Print3Smallest(arr, n): firstmin = MAX secmin = MAX thirdmin = MAX for i in range(0, n): # Check if current element # is less than firstmin, # then update first,second # and third if arr[i] < firstmin: thirdmin = secmin secmin = firstmin firstmin = arr[i] # Check if current element is # less than secmin then update # second and third elif arr[i] < secmin: thirdmin = secmin secmin = arr[i] # Check if current element is # less than,then upadte third elif arr[i] < thirdmin: thirdmin = arr[i] print("First min = ", firstmin) print("Second min = ", secmin) print("Third min = ", thirdmin) # driver program arr = [4, 9, 1, 32, 12] n = len(arr) Print3Smallest(arr, n) # This code is contributed by Shrikant13\. ``` ## C# ```cs // C# program to find the first, second // and third minimum element in an array using System; class GFG { static void Print3Smallest(int []array, int n) { int firstmin = int.MaxValue; int secmin = int.MaxValue; int thirdmin = int.MaxValue; for (int i = 0; i < n; i++) { /* Check if current element is less than firstmin, then update first, second and third */ if (array[i] < firstmin) { thirdmin = secmin; secmin = firstmin; firstmin = array[i]; } /* Check if current element is less than secmin then update second and third */ else if (array[i] < secmin) { thirdmin = secmin; secmin = array[i]; } /* Check if current element is less than then update third */ else if (array[i] < thirdmin) thirdmin = array[i]; } Console.WriteLine("First min = " + firstmin ); Console.WriteLine("Second min = " + secmin ); Console.WriteLine("Third min = " + thirdmin ); } // Driver code static void Main() { int []array = new int[]{4, 9, 1, 32, 12}; int n = array.Length; Print3Smallest(array, n); } } // This code is contributed by Sam007 ``` ## PHP ```php <?php // php program to find the first, second // and third minimum element in an array function Print3Smallest($array, $n) { $MAX = 100000; $firstmin = $MAX; $secmin = $MAX; $thirdmin = $MAX; for ($i = 0; $i < $n; $i++) { /* Check if current element is less than firstmin, then update first, second and third */ if ($array[$i] < $firstmin) { $thirdmin = $secmin; $secmin = $firstmin; $firstmin = $array[$i]; } /* Check if current element is less than secmin then update second and third */ else if ($array[$i] < $secmin) { $thirdmin = $secmin; $secmin = $array[$i]; } /* Check if current element is less than then update third */ else if ($array[$i] < $thirdmin) $thirdmin = $array[$i]; } echo "First min = ".$firstmin."\n"; echo "Second min = ".$secmin."\n"; echo "Third min = ".$thirdmin."\n"; } // Driver code $array= array(4, 9, 1, 32, 12); $n = sizeof($array) / sizeof($array[0]); Print3Smallest($array, $n); // This code is contributed by mits ?> ``` Output: ``` First min = 1 Second min = 4 Third min = 9 ``` * * * * * *