# 查找矩阵所有行共有的不同元素 > 原文： [https://www.geeksforgeeks.org/find-distinct-elements-common-rows-matrix/](https://www.geeksforgeeks.org/find-distinct-elements-common-rows-matrix/) 给定一个 n x n 矩阵。 问题是要找到矩阵所有行共有的所有不同元素。 元素可以按任何顺序打印。 **示例**： ``` Input : mat[][] = { {2, 1, 4, 3}, {1, 2, 3, 2}, {3, 6, 2, 3}, {5, 2, 5, 3} } Output : 2 3 Input : mat[][] = { {12, 1, 14, 3, 16}, {14, 2, 1, 3, 35}, {14, 1, 14, 3, 11}, {14, 25, 3, 2, 1}, {1, 18, 3, 21, 14} } Output : 1 3 14 ``` **方法 1**：使用三个嵌套循环。 检查第一行的元素是否存在于所有后续行中。 O（n ³ ）的时间复杂度。 可能需要额外的空间来处理重复的元素。 **方法 2**：按升序分别对矩阵的所有行进行排序。 然后对[在 3 个排序的数组](https://www.geeksforgeeks.org/find-common-elements-three-sorted-arrays/)中查找公共元素的问题应用改进的方法。 下面给出了相同的实现。 ## C++ ```cpp // C++ implementation to find distinct elements // common to all rows of a matrix #include <bits/stdc++.h> using namespace std; const int MAX = 100; // function to individually sort // each row in increasing order void sortRows(int mat[][MAX], int n) {     for (int i=0; i<n; i++)         sort(mat[i], mat[i] + n); } // function to find all the common elements void findAndPrintCommonElements(int mat[][MAX], int n) {     // sort rows individually     sortRows(mat, n);     // current column index of each row is stored     // from where the element is being searched in     // that row     int curr_index[n];     memset(curr_index, 0, sizeof(curr_index));     int f = 0;     for (; curr_index[0]<n; curr_index[0]++)     {         // value present at the current column index         // of 1st row         int value = mat[0][curr_index[0]];         bool present = true;         // 'value' is being searched in all the         // subsequent rows         for (int i=1; i<n; i++)         {             // iterate through all the elements of             // the row from its current column index             // till an element greater than the 'value'             // is found or the end of the row is             // encountered             while (curr_index[i] < n &&                    mat[i][curr_index[i]] <= value)                 curr_index[i]++;             // if the element was not present at the column             // before to the 'curr_index' of the row             if (mat[i][curr_index[i]-1] != value)                 present = false;             // if all elements of the row have             // been traversed             if (curr_index[i] == n)             {                 f = 1;                 break;             }         }         // if the 'value' is common to all the rows         if (present)             cout << value << " ";         // if any row have been completely traversed         // then no more common elements can be found         if (f == 1)             break;     } } // Driver program to test above int main() {     int mat[][MAX] = {  {12, 1, 14, 3, 16},         {14, 2, 1, 3, 35},         {14, 1, 14, 3, 11},         {14, 25, 3, 2, 1},         {1, 18, 3, 21, 14}     };     int n = 5;     findAndPrintCommonElements(mat, n);     return 0; } ```