方陣中的最大值和最小值。 · GeeksForGeeks 中文系列教程

# 方陣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/) 給定 n * n 階方陣，從給定的矩陣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 ``` **樸素的方法**：我們使用線性搜索分別找到矩陣的最大值和最小值。所需的比較數是 n 2 用于找到最小值，n 2 用于找到最大值元素。總比較等于 2n 2 。 **對比較（有效方法）**：從矩陣中選擇兩個元素，一個從矩陣行的開頭選擇，另一個從矩陣同一行的結尾選擇，比較它們，然后將其比較小它們最大到矩陣的最小值，最大到最大。我們可以看到，對于兩個元素，我們需要 3 比較，因此對于遍歷整個矩陣，我們總共需要 3/2 n 2 比較。 **注意**：這是[最大數組最小值的方法 3 的擴展形式。](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\. ?> ``` **輸出**： ``` Maximum = 25, Minimum = 0 ``` 本文由 [**Shivam Pradhan（anuj_charm）**](https://www.facebook.com/anuj.charm) 提供。如果您喜歡 GeeksforGeeks 并希望做出貢獻，則還可以使用 [tribution.geeksforgeeks.org](http://www.contribute.geeksforgeeks.org) 撰寫文章，或將您的文章郵寄至 tribution@geeksforgeeks.org。查看您的文章出現在 GeeksforGeeks 主頁上，并幫助其他 Geeks。