本文概述给定一个整数数组, 我们的任务是编写一个程序, 该程序可以有效地找到数组中存在的第二大元素。
例子：

Input: arr[] = {12, 35, 1, 10, 34, 1} Output: The second largest element is 34. Explanation: The largest element of the array is 35 and the second largest element is 34Input: arr[] = {10, 5, 10} Output: The second largest element is 5. Explanation: The largest element of the array is 10 and the second largest element is 5Input: arr[] = {10, 10, 10} Output: The second largest does not exist. Explanation: Largest element of the array is 10 there is no second largest element

推荐：请在"
实践
首先, 在继续解决方案之前。
简单的解决方案
方法：
想法是按降序对数组进行排序, 然后返回第二个元素, 该元素不等于已排序数组中的最大元素。
C ++ 14

// C++ program to find second largest // element in an array #include < bits/stdc++.h> using namespace std; /* Function to print the second largest elements */ void print2largest( int arr[], int arr_size) { int i, first, second; /* There should be atleast two elements */ if (arr_size < 2) { printf ( " Invalid Input " ); return ; } // sort the array sort(arr, arr + arr_size); // start from second last element // as the largest element is at last for (i = arr_size - 2; i > = 0; i--) { // if the element is not // equal to largest element if (arr[i] != arr[arr_size - 1]) { printf ( "The second largest element is %d\n" , arr[i]); return ; } } printf ( "There is no second largest element\n" ); } /* Driver program to test above function */ int main() { int arr[] = { 12, 35, 1, 10, 34, 1 }; int n = sizeof (arr) / sizeof (arr[0]); print2largest(arr, n); return 0; }

Java

// Java program to find second largest // element in an array import java.util.*; class GFG{ // Function to print the // second largest elements static void print2largest( int arr[], int arr_size) { int i, first, second; // There should be // atleast two elements if (arr_size < 2 ) { System.out.printf( " Invalid Input " ); return ; } // Sort the array Arrays.sort(arr); // Start from second last element // as the largest element is at last for (i = arr_size - 2 ; i > = 0 ; i--) { // If the element is not // equal to largest element if (arr[i] != arr[arr_size - 1 ]) { System.out.printf( "The second largest " + "element is %d\n" , arr[i]); return ; } } System.out.printf( "There is no second " + "largest element\n" ); } // Driver code public static void main(String[] args) { int arr[] = { 12 , 35 , 1 , 10 , 34 , 1 }; int n = arr.length; print2largest(arr, n); } } // This code is contributed by gauravrajput1

C#

// C# program to find second largest // element in an array using System; class GFG{ // Function to print the // second largest elements static void print2largest( int []arr, int arr_size) { int i; // There should be // atleast two elements if (arr_size < 2) { Console.Write( " Invalid Input " ); return ; } // Sort the array Array.Sort(arr); // Start from second last element // as the largest element is at last for (i = arr_size - 2; i > = 0; i--) {// If the element is not // equal to largest element if (arr[i] != arr[arr_size - 1]) { Console.Write( "The second largest " + "element is {0}\n" , arr[i]); return ; } } Console.Write( "There is no second " + "largest element\n" ); } // Driver code public static void Main(String[] args) { int []arr = { 12, 35, 1, 10, 34, 1 }; int n = arr.Length; print2largest(arr, n); } } // This code is contributed by Amit Katiyar

输出如下：

The second largest element is 34

复杂度分析：

• 时间复杂度：O(n log n)。
  排序数组所需的时间为O(n log n)。
• 辅助空间：O(1)。
  由于不需要额外的空间。

更好的解决方案：
方法：
方法是遍历数组两次。在第一个遍历中找到最大元素。在第二遍历中找到小于在第一遍历中获得的元素的最大元素。
C ++ 14

// C++ program to find second largest // element in an array #include < bits/stdc++.h> using namespace std; /* Function to print the second largest elements */ void print2largest( int arr[], int arr_size) { int i, first, second; /* There should be atleast two elements */ if (arr_size < 2) { printf ( " Invalid Input " ); return ; } int largest = second = INT_MIN; // find the largest element for ( int i = 0; i < arr_size; i++) { largest = max(largest, arr[i]); } // find the second largest element for ( int i = 0; i < arr_size; i++) { if (arr[i] != largest) second = max(second, arr[i]); } if (second == INT_MIN) printf ( "There is no second largest element\n" ); else printf ( "The second largest element is %d\n" , second); } /* Driver program to test above function */ int main() { int arr[] = { 12, 35, 1, 10, 34, 1 }; int n = sizeof (arr) / sizeof (arr[0]); print2largest(arr, n); return 0; }

Java

// Java program to find second largest // element in an array class GFG{ // Function to print the second largest elements static void print2largest( int arr[], int arr_size) { int i, first, second; // There should be atleast two elements if (arr_size < 2 ) { System.out.printf( " Invalid Input " ); return ; } int largest = second = Integer.MIN_VALUE; // Find the largest element for (i = 0 ; i < arr_size; i++) { largest = Math.max(largest, arr[i]); } // Find the second largest element for (i = 0 ; i < arr_size; i++) { if (arr[i] != largest) second = Math.max(second, arr[i]); } if (second == Integer.MIN_VALUE) System.out.printf( "There is no second " + "largest element\n" ); else System.out.printf( "The second largest " + "element is %d\n" , second); } // Driver code public static void main(String[] args) { int arr[] = { 12 , 35 , 1 , 10 , 34 , 1 }; int n = arr.length; print2largest(arr, n); } } // This code is contributed by Amit Katiyar

Python3

# Python3 program to find # second largest element # in an array # Function to print # second largest elements def print2largest(arr, arr_size): # There should be atleast # two elements if (arr_size < 2 ): print ( " Invalid Input " ); return ; largest = second = - 2454635434 ; # Find the largest element for i in range ( 0 , arr_size): largest = max (largest, arr[i]); # Find the second largest element for i in range ( 0 , arr_size): if (arr[i] ! = largest): second = max (second, arr[i]); if (second = = - 2454635434 ): print ( "There is no second " + "largest element" ); else : print ( "The second largest " + "element is \n" , second); # Driver code if __name__ = = '__main__' :arr = [ 12 , 35 , 1 , 10 , 34 , 1 ]; n = len (arr); print2largest(arr, n); # This code is contributed by shikhasingrajput

C#

// C# program to find second largest // element in an array using System; class GFG{ // Function to print the second largest elements static void print2largest( int []arr, int arr_size) { // int first; int i, second; // There should be atleast two elements if (arr_size < 2) { Console.Write( " Invalid Input " ); return ; } int largest = second = int .MinValue; // Find the largest element for (i = 0; i < arr_size; i++) { largest = Math.Max(largest, arr[i]); } // Find the second largest element for (i = 0; i < arr_size; i++) { if (arr[i] != largest) second = Math.Max(second, arr[i]); }if (second == int .MinValue) Console.Write( "There is no second " + "largest element\n" ); else Console.Write( "The second largest " + "element is {0}\n" , second); } // Driver code public static void Main(String[] args) { int []arr = { 12, 35, 1, 10, 34, 1 }; int n = arr.Length; print2largest(arr, n); } } // This code is contributed by Amit Katiyar

输出如下：

The second largest element is 34

复杂度分析：

• 时间复杂度：上)。
  需要两次遍历数组。
• 辅助空间：O(1)。
  由于不需要额外的空间。

高效的解决方案
方法：
在单个遍历中查找第二大元素。
以下是完成此操作的完整算法：

1) Initialize two variables first and second to INT_MIN as first = second = INT_MIN 2) Start traversing the array, a) If the current element in array say arr[i] is greater than first. Then update first and second as, second = first first = arr[i] b) If the current element is in between first and second, then update second to store the value of current variable as second = arr[i] 3) Return the value stored in second.

C

// C program to find second largest // element in an array #include < limits.h> #include < stdio.h> /* Function to print the second largest elements */ void print2largest( 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_MIN; for (i = 0; i < arr_size; i++) { /* If current element is greater 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_MIN) printf ( "There is no second largest element\n" ); else printf ( "The second largest element is %dn" , second); } /* Driver program to test above function */ int main() { int arr[] = { 12, 35, 1, 10, 34, 1 }; int n = sizeof (arr) / sizeof (arr[0]); print2largest(arr, n); return 0; }

C ++

// C++ program to find second largest // element in an array #include < bits/stdc++.h> using namespace std; // Function to print the second // largest elements void print2largest( 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_MIN; for (i = 0; i < arr_size; i++) {// If current element is greater // 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_MIN) cout < < "There is no second largest" "element\n" ; else cout < < "The second largest element is " < < second; } // Driver code int main() { int arr[] = { 12, 35, 1, 10, 34, 1 }; int n = sizeof (arr) / sizeof (arr[0]); print2largest(arr, n); return 0; } // This code is contributed by shivanisinghss2110

Java

// JAVA Code for Find Second largest // element in an array class GFG { /* Function to print the second largest elements */ public static void print2largest( int arr[], int arr_size) { int i, first, second; /* There should be atleast two elements */ if (arr_size < 2 ) { System.out.print( " Invalid Input " ); return ; } first = second = Integer.MIN_VALUE; 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 == Integer.MIN_VALUE) System.out.print( "There is no second largest" + " element\n" ); else System.out.print( "The second largest element" + " is " + second); } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = { 12 , 35 , 1 , 10 , 34 , 1 }; int n = arr.length; print2largest(arr, n); } } // This code is contributed by Arnav Kr. Mandal.

Python3

# Python program to # find second largest # element in an array # Function to print the # second largest elements def print2largest(arr, arr_size): # There should be atleast # two elements if (arr_size < 2 ):print ( " Invalid Input " ) return first = second = - 2147483648 for i in range (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 = = - 2147483648 ): print ( "There is no second largest element" ) else : print ( "The second largest element is" , second) # Driver program to test # above function arr = [ 12 , 35 , 1 , 10 , 34 , 1 ] n = len (arr) print2largest(arr, n) # This code is contributed # by Anant Agarwal.

C#

// C# Code for Find Second largest // element in an array using System; class GFG { // Function to print the // second largest elements public static void print2largest( int [] arr, int arr_size) { int i, first, second; // There should be atleast two elements if (arr_size < 2) { Console.WriteLine( " Invalid Input " ); return ; } first = second = int .MinValue; 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 .MinValue) Console.Write( "There is no second largest" + " element\n" ); else Console.Write( "The second largest element" + " is " + second); } // Driver Code public static void Main(String[] args) { int [] arr = { 12, 35, 1, 10, 34, 1 }; int n = arr.Length; print2largest(arr, n); } } // This code is contributed by Parashar.

的PHP

< ?php // PHP program to find second largest // element in an array // Function to print the // second largest elements function print2largest( $arr , $arr_size ) { // There should be atleast // two elements if ( $arr_size < 2) { echo ( " Invalid Input " ); return ; } $first = $second = PHP_INT_MIN; 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 == PHP_INT_MIN) echo ( "There is no second largest element\n" ); else echo ( "The second largest element is " . $second . "\n" ); } // Driver Code $arr = array (12, 35, 1, 10, 34, 1); $n = sizeof( $arr ); print2largest( $arr , $n ); // This code is contributed by Ajit. ?>

输出如下：

The second largest element is 34

【在数组中查找第二大元素（详细代码实现）】复杂度分析：

• 时间复杂度：上)。
  只需遍历数组。
• 辅助空间：O(1)。
  由于不需要额外的空间。

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