# 计算矩阵中所有排序的行 > 原文： [https://www.geeksforgeeks.org/count-sorted-rows-matrix/](https://www.geeksforgeeks.org/count-sorted-rows-matrix/) 给定一个 m * n 大小的矩阵，任务是对矩阵中按严格递增顺序或严格递减顺序排序的所有行进行计数吗？ **示例**： ``` Input : m = 4, n = 5 mat[m][n] = 1 2 3 4 5 4 3 1 2 6 8 7 6 5 4 5 7 8 9 10 Output: 3 ``` 这个想法很简单，涉及矩阵的两个遍历。 1）从矩阵左侧遍历以严格按**顺序递增的所有行** 2）从矩阵右侧遍历以对[[ **严格降序** 以下是上述想法的实现。 ## C++ ```cpp // C++ program to find number of sorted rows #include <bits/stdc++.h> #define MAX 100 using namespace std; // Function to count all sorted rows in a matrix int sortedCount(int mat[][MAX], int r, int c) {     int result = 0; // Initialize result     // Start from left side of matrix to     // count increasing order rows     for (int i=0; i<r; i++)     {         // Check if there is any pair ofs element         // that are  not in increasing order.         int j;         for (j=0; j<c-1; j++)             if (mat[i][j+1] <= mat[i][j])                 break;         // If the loop didn't break (All elements         // of current row were in increasing order)         if (j == c-1)             result++;     }     // Start from right side of matrix to     // count increasing order rows ( reference     // to left these are in decreasing order )     for (int i=0; i<r; i++)     {         // Check if there is any pair ofs element         // that are  not in decreasing order.         int j;         for (j=c-1; j>0; j--)             if (mat[i][j-1] <= mat[i][j])                 break;         // Note c > 1 condition is required to make         // sure that a single column row is not counted         // twice (Note that a single column row is sorted         // both in increasing and decreasing order)          if (c > 1 && j == 0)             result++;     }     return result; } // Driver program to run the case int main() {     int m = 4, n = 5;     int mat[][MAX] = {{1, 2, 3, 4, 5},                       {4, 3, 1, 2, 6},                       {8, 7, 6, 5, 4},                       {5, 7, 8, 9, 10}};     cout << sortedCount(mat, m, n);     return 0; } ```