# 以对角线图案打印矩阵 > 原文： [https://www.geeksforgeeks.org/print-matrix-diagonal-pattern/](https://www.geeksforgeeks.org/print-matrix-diagonal-pattern/) 给定 n * n 大小的矩阵，任务是以对角线图案打印其元素。  ``` Input : mat[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}} Output : 1 2 4 7 5 3 6 8 9. Explanation: Start from 1 Then from upward to downward diagonally i.e. 2 and 4 Then from downward to upward diagonally i.e 7, 5, 3 Then from up to down diagonally i.e 6, 8 Then down to up i.e. end at 9. Input : mat[4][4] = {{1, 2, 3, 10}, {4, 5, 6, 11}, {7, 8, 9, 12}, {13, 14, 15, 16}} Output: 1 2 4 7 5 3 10 6 8 13 14 9 11 12 15 16 . Explanation: Start from 1 Then from upward to downward diagonally i.e. 2 and 4 Then from downward to upward diagonally i.e 7, 5, 3 Then from upward to downward diagonally i.e. 10 6 8 13 Then from downward to upward diagonally i.e 14 9 11 Then from upward to downward diagonally i.e. 12 15 then end at 16 ``` **方法**：从图中可以看出，每个元素要么斜向上打印，要么斜向下打印。 从索引（0,0）开始，对角向上打印元素，然后更改方向，改变列，对角向下打印。 该循环一直持续到到达最后一个元素为止。 **算法**： 1. 创建变量 *i = 0，j = 0* 以存储行和列的当前索引 2. 运行从 0 到 n * n 的循环，其中 n 是矩阵的边。 3. 使用标志 **isUp** 来确定方向是向上还是向下。 最初将 *isUp = true* 设置为向上方向。 4. 如果 isUp = 1，则通过增加列索引和减少行索引来开始打印元素。 5. 同样，如果 isUp = 0，则减少列索引并增加行索引。 6. 移至下一个列或行（下一个开始的行和列） 7. 这样做直到遍历所有元素。 **实现**： ## C++ ```cpp // C++ program to print matrix in diagonal order #include <bits/stdc++.h> using namespace std; const int MAX = 100; void printMatrixDiagonal(int mat[MAX][MAX], int n) {     // Initialize indexes of element to be printed next     int i = 0, j = 0;     // Direction is initially from down to up     bool isUp = true;     // Traverse the matrix till all elements get traversed     for (int k = 0; k < n * n;) {         // If isUp = true then traverse from downward         // to upward         if (isUp) {             for (; i >= 0 && j < n; j++, i--) {                 cout << mat[i][j] << " ";                 k++;             }             // Set i and j according to direction             if (i < 0 && j <= n - 1)                 i = 0;             if (j == n)                 i = i + 2, j--;         }         // If isUp = 0 then traverse up to down         else {             for (; j >= 0 && i < n; i++, j--) {                 cout << mat[i][j] << " ";                 k++;             }             // Set i and j according to direction             if (j < 0 && i <= n - 1)                 j = 0;             if (i == n)                 j = j + 2, i--;         }         // Revert the isUp to change the direction         isUp = !isUp;     } } int main() {     int mat[MAX][MAX] = { { 1, 2, 3 },                           { 4, 5, 6 },                           { 7, 8, 9 } };     int n = 3;     printMatrixDiagonal(mat, n);     return 0; } ```