# 给定 1、2、3……k 以之字形打印它们。 > 原文： [https://www.geeksforgeeks.org/given-1s-2s-3s-ks-print-zig-zag-way/](https://www.geeksforgeeks.org/given-1s-2s-3s-ks-print-zig-zag-way/) 给定的行数和列数。 并给定需要打印的数字 1、2、3…k。 以之字形打印它们。 确保 n * m = 1 的个数+ 2 的个数+ 3 的+…+ k 的个数 例子： ``` Input : 2 3 2 1 2 1 Output : 1 1 2 4 3 3 Explanation : Here number of rows are 2 and number of columns are 3 and number of 1's are 2 number of 2's are 1 number of 3's are 2 number of 4's are 1 ----------- | 1 | 1 | 2 | | 3 | 3 | 4 | ----------- Input : 4 3 2 4 3 1 2 Output : 1 1 2 2 2 2 3 3 3 5 5 4 Explanation : Here number of rows are 4 and number of columns are 3 and number of 1's are 2 number of 2's are 4 [Note that 2s are printed in] number of 3's are 3 [zig zag manner] number of 4's are 1 number of 5's are 2 ``` **方法**：我们制作了一个二维数组，以之字形方式存储所有元素。 我们将遍历数字数组的所有元素，并将第 i 个索引的数组的所有数字插入二维数组，直到其变为零。 ## C++ ```cpp // CPP program to print  given number of 1's,  // 2's, 3's ....k's in  zig-zag way. #include <bits/stdc++.h> using namespace std; // function that prints  given number of 1's,  // 2's, 3's ....k's in zig-zag way. void ZigZag(int rows, int columns, int numbers[]) {     int k = 0;     // two-dimensional array to store numbers.      int arr[rows][columns];     for (int i=0; i<rows; i++)     {         // for even row.         if (i%2==0)         {             // for each column.             for (int j=0; j<columns and                      numbers[k]>0; j++)             {                 // storing element.                 arr[i][j] = k+1;                 // decrement element at                  // kth index.                  numbers[k]--;                 // if array contains zero                 // then increment index to                  // make this next index                 if (numbers[k] == 0)                     k++;             }         }         // for odd row.         else         {             // for each column.             for (int j=columns-1; j>=0 and                  numbers[k]>0; j--)             {                 // storing element.                 arr[i][j] = k+1;                 // decrement element                  // at kth index.                 numbers[k]--;                 // if array contains zero then                  // increment index to make this                  // next index.                 if (numbers[k]==0)                     k++;             }         }     }     // printing the stored elements.     for (int i=0;i<rows;i++)     {         for (int j=0;j<columns;j++)             cout << arr[i][j] << " ";         cout << endl;     } } // Driver code for above function. int main() {     int rows = 4;     int columns = 5;     int Numbers[] = {3, 4, 2, 2, 3, 1, 5};     ZigZag(rows, columns, Numbers);         return 0; }  ```