排序给定矩阵 · GeeksForGeeks 中文系列教程

# 排序给定矩阵 > 原文： [https://www.geeksforgeeks.org/sort-given-matrix/](https://www.geeksforgeeks.org/sort-given-matrix/) 给定一个 n x n 矩阵。问题是要对给定矩阵进行严格排序。在这里，严格顺序表示矩阵的排序方式是，一行中的所有元素都以升序排序，并且对于行'i'，其中 1 < = i < = n-1，行'的第一个元素 i'大于或等于行'i-1'的最后一个元素。例子： ``` Input : mat[][] = { {5, 4, 7}, {1, 3, 8}, {2, 9, 6} } Output : 1 2 3 4 5 6 7 8 9 ``` **方法**：创建大小为 n ^ 2 的 **temp []** 数组。从第一行开始，一个一个地复制给定矩阵的元素到 temp []中。排序 temp []。现在，将 temp []的元素一一复制到给定的矩阵中。 ## C++ ```cpp // C++ implementation to sort the given matrix #include <bits/stdc++.h> using namespace std; #define SIZE 10 // function to sort the given matrix void sortMat(int mat[SIZE][SIZE], int n) { // temporary matrix of size n^2 int temp[n * n]; int k = 0; // copy the elements of matrix one by one // into temp[] for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) temp[k++] = mat[i][j]; // sort temp[] sort(temp, temp + k); // copy the elements of temp[] one by one // in mat[][] k = 0; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) mat[i][j] = temp[k++]; } // function to print the given matrix void printMat(int mat[SIZE][SIZE], int n) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) cout << mat[i][j] << " "; cout << endl; } } // Driver program to test above int main() { int mat[SIZE][SIZE] = { { 5, 4, 7 }, { 1, 3, 8 }, { 2, 9, 6 } }; int n = 3; cout << "Original Matrix:\n"; printMat(mat, n); sortMat(mat, n); cout << "\nMatrix After Sorting:\n"; printMat(mat, n); return 0; } ``` ## Java ```java // Java implementation to // sort the given matrix import java.io.*; import java.util.*; class GFG { static int SIZE = 10; // function to sort the given matrix static void sortMat(int mat[][], int n) { // temporary matrix of size n^2 int temp[] = new int[n * n]; int k = 0; // copy the elements of matrix // one by one into temp[] for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) temp[k++] = mat[i][j]; // sort temp[] Arrays.sort(temp); // copy the elements of temp[] // one by one in mat[][] k = 0; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) mat[i][j] = temp[k++]; } // function to print the given matrix static void printMat(int mat[][], int n) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) System.out.print( mat[i][j] + " "); System.out.println(); } } // Driver program to test above public static void main(String args[]) { int mat[][] = { { 5, 4, 7 }, { 1, 3, 8 }, { 2, 9, 6 } }; int n = 3; System.out.println("Original Matrix:"); printMat(mat, n); sortMat(mat, n); System.out.println("Matrix After Sorting:"); printMat(mat, n); } } // This code is contributed by Nikita Tiwari. ``` ## Python3 ```py # Python3 implementation to sort # the given matrix SIZE = 10 # Function to sort the given matrix def sortMat(mat, n) : # Temporary matrix of size n^2 temp = [0] * (n * n) k = 0 # Copy the elements of matrix # one by one into temp[] for i in range(0, n) : for j in range(0, n) : temp[k] = mat[i][j] k += 1 # sort temp[] temp.sort() # copy the elements of temp[] # one by one in mat[][] k = 0 for i in range(0, n) : for j in range(0, n) : mat[i][j] = temp[k] k += 1 # Function to print the given matrix def printMat(mat, n) : for i in range(0, n) : for j in range( 0, n ) : print(mat[i][j] , end = " ") print() # Driver program to test above mat = [ [ 5, 4, 7 ], [ 1, 3, 8 ], [ 2, 9, 6 ] ] n = 3 print( "Original Matrix:") printMat(mat, n) sortMat(mat, n) print("\nMatrix After Sorting:") printMat(mat, n) # This code is contributed by Nikita Tiwari. ``` ## C# ```cs // C# implementation to // sort the given matrix using System; class GFG { static int SIZE = 10; // function to sort the given matrix static void sortMat(int[, ] mat, int n) { // temporary matrix of size n^2 int[] temp = new int[n * n]; int k = 0; // copy the elements of matrix // one by one into temp[] for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) temp[k++] = mat[i, j]; // sort temp[] Array.Sort(temp); // copy the elements of temp[] // one by one in mat[][] k = 0; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) mat[i, j] = temp[k++]; } // function to print the given matrix static void printMat(int[, ] mat, int n) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) Console.Write(mat[i, j] + " "); Console.WriteLine(); } } // Driver code public static void Main() { int[, ] mat = { { 5, 4, 7 }, { 1, 3, 8 }, { 2, 9, 6 } }; int n = 3; Console.WriteLine("Original Matrix:"); printMat(mat, n); sortMat(mat, n); Console.WriteLine("Matrix After Sorting:"); printMat(mat, n); } } // This code is contributed by Sam007 ``` Output : ``` Original Matrix: 5 4 7 1 3 8 2 9 6 Matrix After Sorting: 1 2 3 4 5 6 7 8 9 ``` 时间复杂度：O（n <sup>2</sup> log <sub>2</sub> n）。辅助空间：`O(n^2)`。 * * * * * *