我试图用Java创建一个键盘,用来输入用户输入的字母,以及每行所需的字符数。然后应该以所需的行数打印字符,因此如果输入"abcdefgh"
并且所需的行号为4,则应打印:
abcd
efgh
但我仍然坚持如何让它发挥作用。
public class Keypad {
char [][] letters;
public Keypad(String chars, int rowLength) {
int counter = 0;
for (int i = 0; i<chars.length(); i++){
counter++;
}
letters = new char[rowLength][counter/rowLength];
}
public String toString() {
String s = " ";
for (int row=0; row<letters.length; row=row+1) { // Over rows
for (int col=0; col<letters[row].length; col=col+1) {
s = s + letters[row][col];
}
s = s + "\n";
}
return "the keypad is" + s;
}
答案 0 :(得分:3)
toString()
方法的逻辑看起来很好,但是你没有在构造函数中填充letters
数组。所以你需要在构造函数中添加这样的东西:
public Keypad(String chars, int rowLength) {
// you don't need to count the length with a loop
int nRow = chars.length()/rowLength;
if(chars.length()%rowLength!=0) nRow++;
letters = new char[nRow][rowLength];
for(int i = 0, n = 0 ; i < letters.length ; i++) {
for(int j = 0 ; n < chars.length() && j < letters[i].length ; j++, n++) {
letters[i][j] = chars.charAt(n);
}
}
}