[英]Printing pattern using Jagged Array
给定锯齿状数组,我们被要求使用循环语句来显示基于 position 的字符。如果 position 匹配则显示“*”,否则显示“”。
int arr [][] = {{0,4,8,12,13,14,15,18,19,20,21,24,28},
{0,4,7,9,12,16,18,22,25,27},
{0,1,2,3,4,6,10,12,16,18,22,26},
{0,4,6,10,12,13,14,15,18,19,20,21,26},
{0,4,6,7,8,9,10,12,18,26},
{0,4,6,10,12,18,26}};
我已经创建了一个程序,但是 output 不是我所期望的,我现在卡住了。
for (int i = 0; i < arr.length; i++)
{
for (int j = 0; j < arr[i].length - 1; j++)
{
for (int spaces = 1; spaces < arr[i][j + 1]-arr[i][j]; spaces++)
{
System.out.print(" ");
}
System.out.print("*");
}
System.out.println();
}
output 应该是 Happy,但我得到:在此处输入图片描述
这是因为您的程序没有将第一个j
值(即 0)与自身进行比较。 由于每个值都等于自身,因此您可以像这样为每一行手动添加*
。
int arr [][] = {{0,4,8,12,13,14,15,18,19,20,21,24,28},
{0,4,7,9,12,16,18,22,25,27},
{0,1,2,3,4,6,10,12,16,18,22,26},
{0,4,6,10,12,13,14,15,18,19,20,21,26},
{0,4,6,7,8,9,10,12,18,26},
{0,4,6,10,12,18,26}};
for (int i = 0; i < arr.length; i++)
{
System.out.print("*");
for (int j = 0; j < arr[i].length - 1; j++)
{
for (int spaces = 1; spaces < arr[i][j + 1]-arr[i][j]; spaces++)
{
System.out.print(" ");
}
System.out.print("*");
}
System.out.println();
}
尝试这个。
for (int i = 0; i < arr.length; i++) {
for (int j = 0, p = -1; j < arr[i].length; p = arr[i][j++])
System.out.print(" ".repeat(arr[i][j] - p - 1) + "*");
System.out.println();
}
output:
* * * **** **** * *
* * * * * * * * * *
***** * * * * * * *
* * * * **** **** *
* * ***** * * *
* * * * * * *
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