我正在尝试将用户制作的数组(大小在3-11之间)填充到某个元素位置的字符以获取模式。用户输入的内容同时充当行数和列数,因此,如果他们在下面的示例中输入5,则他们将获得5乘5的数组。我想要获得这种模式
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* * * * *
* * *
*
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* * * * *
* * * * *
* * * * *
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这是代码
public static void main (String [] args) {
int dimension = findDimension();
char [] [] array2d = new char [dimension] [dimension];
char star = '*';
array2d = pointDown(star,dimension);
System.out.println();
print(array2d);
}
public static void print(char [] [] arrayParam) {
for (int hyphen = 0; hyphen < (arrayParam.length*2)+1; hyphen++) {
System.out.print("-");
}
System.out.println();
for(char[] row : arrayParam)
{
for(char c : row)
System.out.print(" " + c);
System.out.printf("\n");
}
for (int hyphen = 0; hyphen < (arrayParam.length*2)+1; hyphen++) {
System.out.print("-");
}
}
问题应该在于这个方法,我认为这个循环之后
public static char [] [] pointDown (char starParam, int dimenParam) {
char [] [] pointDown = new char [dimenParam] [dimenParam];
for (int i = 0; i < dimenParam; i++){
for (int j = 0; j < dimenParam; j++) {
pointDown[i][j] = ' ';
// I fill the positions first with blank spaces then add the characters
// with the loop below
}
}
/* Problem should be in this loop, Is there even a pattern to it though
* since columns would have to account for both the last and beginning
* columns after the first loop? Should I make variables for those or is
*/ there a simpler way to do it that I'm missing?
for (int i = 0; i <= dimenParam/2; i++) {
for (int j = 0; j < dimenParam; j++) {
pointDown[i][j] = starParam;
}
}
return pointDown;
}
更新:在考虑到我被告知的内容后,我能够找出问题所在。这是代码应该是什么样子
char [] [] pointDown = new char [dimenParam] [dimenParam];
for (int i = 0; i < dimenParam; i++){
for (int j = 0; j < dimenParam; j++) {
pointDown[i][j] = ' ';
// As before this part fills the array with blank spaces
}
}
int columnEnd = dimenParam; // Set up a variable to hold how far the column goes to
int j = 0;
for (int row = 0; row <= dimenParam/2; row++) {
for (int column = j; column < columnEnd; column++) {
pointDown[row][column] = starParam;
}
columnEnd--; // I had to decrease the ending column in the outer loop
j++; // Originally I had this in the inner loop for the longest time
// By moving it to the outer loop I avoid out of Bounds and runtime errors
}
return pointDown;
我很确定我仍然会处理复杂的事情,但我很满意这段代码