数据结构与算法分析Java版练习1.3

package ch01;
/**
 * 练习1.3 只使用处理I/O的printDigit方法，编写一种方法以输出任意
 *         double型量(可以是负数）。
 */
public class EX03 {
private static void printDigit(int i) {
System.out.print(i);
}

private static void printOut(int n) {
if (n >= 10) {
printOut(n/10);
}
printDigit(n%10);
}

private static void printDouble(double f) {
//首先找出负号
if (f < 0) {
System.out.print('-');
f = -f;
}

//找出小数点位置
String value = String.valueOf(f);
int intPart = 0, dotPart = 0;
int dotPos = value.indexOf('.');
if (dotPos < 0) {
intPart = Integer.valueOf(value);
printOut(intPart);
} else if (dotPos == 0) {
intPart = 0;
dotPart = Integer.valueOf(value.substring(dotPos+1));
printOut(intPart);
System.out.print('.');
printOut(dotPart);
} else {
intPart = Integer.valueOf(value.substring(0, dotPos));
dotPart = Integer.valueOf(value.substring(dotPos + 1));
printOut(intPart);
System.out.print('.');
printOut(dotPart);
} 
System.out.println();
}

public static void main(String[] args) {
printDouble(10.234);
printDouble(-10000.2);
printDouble(1000);
printDouble(10001.0);
printDouble(10001.1456);
}
}