给定 1's, 2's, 3's ......k's 以 zig zag 方式打印它们。
给定的行数和列数。并给出需要打印的 1、2、3 ……k 的数量。以之字形方式打印它们。
保证 n*m = 1 的个数 + 2 的个数 + 3 的个数 + …… + 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：我们制作一个二维数组，以之字形方式存储所有元素。我们将遍历一个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一个二维数组，直到它变为零。
C++
// CPP program to print given number of 1's,
// 2's, 3's ....k's in zig-zag way.
#include
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; i0; 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 Java
// Java program to print given
// number of 1's, 2's, 3's ....k's
// in zig-zag way.
import java.util.*;
import java.lang.*;
public class GfG{
// function that prints given number of 1's,
// 2's, 3's ....k's in zig-zag way.
public static void ZigZag(int rows,
int columns,
int numbers[])
{
int k = 0;
// two-dimensional array to store numbers.
int[][] arr = new int[rows][columns];
for (int i=0; i0; 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 &&
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 Python 3
# Python3 program to print given number of 1's,
# 2's, 3's ....k's in zig-zag way.
# function that prints given number of 1's,
# 2's, 3's ....k's in zig-zag way.
def ZigZag(rows, columns, numbers):
k = 0
# two-dimensional array to store numbers.
arr = [[0 for i in range(columns)] for j in range(rows)]
for i in range(rows):
# for even row.
if (i % 2 == 0):
# for each column.
j = 0
while j < columns and numbers[k] > 0:
# storing element.
arr[i][j] = k + 1
# decrement element at
# kth index.
numbers[k] -= 1
# if array contains zero
# then increment index to
# make this next index
if numbers[k] == 0:
k += 1
j += 1
# for odd row.
else:
# for each column.
j = columns-1
while j>=0 and numbers[k]>0:
# storing element.
arr[i][j] = k+1
# decrement element
# at kth index.
numbers[k] -= 1
# if array contains zero then
# increment index to make this
# next index.
if numbers[k] == 0:
k += 1
j -= 1
# printing the stored elements.
for i in arr:
for j in i:
print(j, end =" ")
print()
# Driver code
rows = 4;
columns = 5;
Numbers = [3, 4, 2, 2, 3, 1, 5]
ZigZag(rows, columns, Numbers)
# This code is contributed by
# Rajnis09C#
// C# program to print given
// number of 1's, 2's, 3's ....k's
// in zig-zag way.
using System;
public class GfG{
// function that prints given number of 1's,
// 2's, 3's ....k's in zig-zag way.
public static void ZigZag(int rows,
int columns,
int []numbers)
{
int k = 0;
// two-dimensional array to store numbers.
int[,] arr = new int[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 &&
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 &&
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++)
Console.Write(arr[i, j] + " ");
Console.WriteLine();
}
}
// Driver function
public static void Main()
{
int rows = 4;
int columns = 5;
int []Numbers = new int[]{3, 4, 2,
2, 3, 1, 5};
ZigZag(rows, columns, Numbers);
}
}
/* This code is contributed by vt_m*/PHP
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)
$k++;
}
}
// for odd row.
else
{
// for each column.
for($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)
$k++;
}
}
}
// printing the stored elements.
for($i = 0; $i < $rows;$i++)
{
for($j = 0; $j < $columns; $j++)
echo $arr[$i][$j] , " ";
echo "\n";
}
}
// Driver Code
$rows = 4;
$columns = 5;
$Numbers = array(3, 4, 2, 2, 3, 1, 5);
ZigZag($rows, $columns,$Numbers);
// This code is contributed by anuj_67.
?>Javascript
输出：
1 1 1 2 2
4 3 3 2 2
4 5 5 5 6
7 7 7 7 7