方阵中的最大值和最小值。 · GeeksForGeeks 中文系列教程

# 方阵中的最大值和最小值。 > 原文： [https://www.geeksforgeeks.org/maximum-and-minimum-in-a-square-matrix/](https://www.geeksforgeeks.org/maximum-and-minimum-in-a-square-matrix/) 给定 n * n 阶方阵，从给定的矩阵中找到最大值和最小值。 **示例**： ``` Input : arr[][] = {5, 4, 9, 2, 0, 6, 3, 1, 8}; Output : Maximum = 9, Minimum = 0 Input : arr[][] = {-5, 3, 2, 4}; Output : Maximum = 4, Minimum = -5 ``` **朴素的方法**：我们使用线性搜索分别找到矩阵的最大值和最小值。所需的比较数是 n 2 用于找到最小值，n 2 用于找到最大值元素。总比较等于 2n 2 。 **对比较（有效方法）**：从矩阵中选择两个元素，一个从矩阵行的开头选择，另一个从矩阵同一行的结尾选择，比较它们，然后将其比较小它们最大到矩阵的最小值，最大到最大。我们可以看到，对于两个元素，我们需要 3 比较，因此对于遍历整个矩阵，我们总共需要 3/2 n 2 比较。 **注意**：这是[最大数组最小值的方法 3 的扩展形式。](https://www.geeksforgeeks.org/maximum-and-minimum-in-an-array/) ## C++ ```cpp // C++ program for finding maximum and minimum in // a matrix. #include<bits/stdc++.h> using namespace std; #define MAX 100 // Finds maximum and minimum in arr[0..n-1][0..n-1] // using pair wise comparisons void maxMin(int arr[][MAX], int n) { int min = INT_MAX; int max = INT_MIN; // Traverses rows one by one for (int i = 0; i < n; i++) { for (int j = 0; j <= n/2; j++) { // Compare elements from beginning // and end of current row if (arr[i][j] > arr[i][n-j-1]) { if (min > arr[i][n-j-1]) min = arr[i][n-j-1]; if (max< arr[i][j]) max = arr[i][j]; } else { if (min > arr[i][j]) min = arr[i][j]; if (max< arr[i][n-j-1]) max = arr[i][n-j-1]; } } } cout << "Maximum = " << max; << ", Minimum = " << min; } /* Driver program to test above function */ int main() { int arr[MAX][MAX] = {5, 9, 11, 25, 0, 14, 21, 6, 4}; maxMin(arr, 3); return 0; } ``` ## Java ```java // Java program for finding maximum // and minimum in a matrix. class GFG { static final int MAX = 100; // Finds maximum and minimum // in arr[0..n-1][0..n-1] // using pair wise comparisons static void maxMin(int arr[][], int n) { int min = +2147483647; int max = -2147483648; // Traverses rows one by one for (int i = 0; i < n; i++) { for (int j = 0; j <= n/2; j++) { // Compare elements from beginning // and end of current row if (arr[i][j] > arr[i][n - j - 1]) { if (min > arr[i][n - j - 1]) min = arr[i][n - j - 1]; if (max< arr[i][j]) max = arr[i][j]; } else { if (min > arr[i][j]) min = arr[i][j]; if (max< arr[i][n - j - 1]) max = arr[i][n - j - 1]; } } } System.out.print("Maximum = "+max+ ", Minimum = "+min); } // Driver program public static void main (String[] args) { int arr[][] = {{5, 9, 11}, {25, 0, 14}, {21, 6, 4}}; maxMin(arr, 3); } } // This code is contributed by Anant Agarwal. ``` ## Python3 ```py # Python3 program for finding # MAXimum and MINimum in a matrix. MAX = 100 # Finds MAXimum and MINimum in arr[0..n-1][0..n-1] # using pair wise comparisons def MAXMIN(arr, n): MIN = 10**9 MAX = -10**9 # Traverses rows one by one for i in range(n): for j in range(n // 2 + 1): # Compare elements from beginning # and end of current row if (arr[i][j] > arr[i][n - j - 1]): if (MIN > arr[i][n - j - 1]): MIN = arr[i][n - j - 1] if (MAX< arr[i][j]): MAX = arr[i][j] else: if (MIN > arr[i][j]): MIN = arr[i][j] if (MAX< arr[i][n - j - 1]): MAX = arr[i][n - j - 1] print("MAXimum =", MAX, ", MINimum =", MIN) # Driver Code arr = [[5, 9, 11], [25, 0, 14], [21, 6, 4]] MAXMIN(arr, 3) # This code is contributed by Mohit Kumar ``` ## C# ```cs // C# program for finding maximum // and minimum in a matrix. using System; public class GFG { // Finds maximum and minimum // in arr[0..n-1][0..n-1] // using pair wise comparisons static void maxMin(int[,] arr, int n) { int min = +2147483647; int max = -2147483648; // Traverses rows one by one for (int i = 0; i < n; i++) { for (int j = 0; j <= n/2; j++) { // Compare elements from beginning // and end of current row if (arr[i,j] > arr[i,n - j - 1]) { if (min > arr[i,n - j - 1]) min = arr[i,n - j - 1]; if (max < arr[i,j]) max = arr[i,j]; } else { if (min > arr[i,j]) min = arr[i,j]; if (max < arr[i,n - j - 1]) max = arr[i,n - j - 1]; } } } Console.Write("Maximum = " + max + ", Minimum = " + min); } // Driver code static public void Main () { int[,] arr = { {5, 9, 11}, {25, 0, 14}, {21, 6, 4} }; maxMin(arr, 3); } } // This code is contributed by Shrikant13\. ``` ## PHP ```php <?php // PHP program for finding // maximum and minimum in // a matrix. $MAX = 100; // Finds maximum and minimum // in arr[0..n-1][0..n-1] // using pair wise comparisons function maxMin($arr, $n) { $min = PHP_INT_MAX; $max = PHP_INT_MIN; // Traverses rows one by one for ($i = 0; $i < $n; $i++) { for ($j = 0; $j <= $n / 2; $j++) { // Compare elements from beginning // and end of current row if ($arr[$i][$j] > $arr[$i][$n - $j - 1]) { if ($min > $arr[$i][$n - $j - 1]) $min = $arr[$i][$n - $j - 1]; if ($max< $arr[$i][$j]) $max = $arr[$i][$j]; } else { if ($min > $arr[$i][$j]) $min = $arr[$i][$j]; if ($max < $arr[$i][$n - $j - 1]) $max = $arr[$i][$n - $j - 1]; } } } echo "Maximum = " , $max ,", Minimum = " , $min; } // Driver Code $arr = array(array(5, 9, 11), array(25, 0, 14), array(21, 6, 4)); maxMin($arr, 3); // This code is contributed by anuj_67\. ?> ``` **输出**： ``` Maximum = 25, Minimum = 0 ``` 本文由 [**Shivam Pradhan（anuj_charm）**](https://www.facebook.com/anuj.charm) 提供。如果您喜欢 GeeksforGeeks 并希望做出贡献，则还可以使用 [tribution.geeksforgeeks.org](http://www.contribute.geeksforgeeks.org) 撰写文章，或将您的文章邮寄至 tribution@geeksforgeeks.org。查看您的文章出现在 GeeksforGeeks 主页上，并帮助其他 Geeks。