Question

java求助一个二进制转ASCII码问题

public static void main(String[] args) {
System.out.println("abcdef".substring(2));
System.out.println(binToAscii("100100001110010000011001000001"));
}

static String binToAscii(String bin) {
return bin.length() > 31 ? bta(bin) : bta(Integer.parseInt(bin, 2));
}

static String bta(int ib) {

String ret = "";
while(ib != 0) {
// 低8位转char
ret = (char) (ib & 0xff) + ret;
// 右移10位，8位+10，高位补0
ib = ib >>> 10;
}
return ret;
}
static String bta(String bin) {
String ret = "";
while(bin.length() >= 10) {
String obin = bin.substring(bin.length() - 8, bin.length());
ret = (char)Integer.parseInt(obin, 2) + ret;
bin = bin.substring(0, bin.length() - 10);
}
if(bin.length() != 0) {
ret = (char)Integer.parseInt(bin, 2) + ret;
}
return ret;
}
上面代码binToAscii("100100001110010000011001000001")运行结果是CAA；是对的；如果是这个的怎么办000001001000011001001000001100100000100000 就是在原来的突然多了几个0怎么解决；想要的结果还是CAA；谢谢各位；求高手；谢谢

Answer 1

帮你改写一下bta函数

static String bta(String bin) {
//加上以下三句，把前面的0去掉
int startIndex = 0;
for (int i = 0; i < bin.length() && bin.charAt(i) == '0'; i++, startIndex++)
;
bin = bin.substring(startIndex);

String ret = "";
while (bin.length() >= 10) {
String obin = bin.substring(bin.length() - 8, bin.length());
ret = (char) Integer.parseInt(obin, 2) + ret;
bin = bin.substring(0, bin.length() - 10);
}
if (bin.length() != 0) {
ret = (char) Integer.parseInt(bin, 2) + ret;
}
return ret;
}

追问

000001001000011000001001000001100100000100000
哥们把前面的0不行呀；就算把的0去掉也不行呀；比如说10C（01000011）与10A之间加几个0就不行了；哥们你看看这个000001001000011000001001000001100100000100000这么搞成CAA；就是先判断一个10；如果有10就开始把后面的数据转码；之后再判断10后面的数据；