java中二进制字符串的Xor

Question

我使用以下代码XOR 2字符串

String lseq = "0011111111101111111111111100101101111100110000001011111000010100";
String tseq = "0011111111100000110011001100110011001100110011001100110011001100";

StringBuilder sb = new StringBuilder();

for(int i = 0; i < lseq.length(); i++)
    sb.append((lseq.charAt(i) ^ tseq.charAt(i)));

String result = sb.toString();
System.out.println(result);

上面的代码给出了正确的输出:

0000000000001111001100110000011110110000000011000111001011011000

我需要再多一个字符串

String hseq = "0011111111110010010111110100010111100000101101001110000100011110";

如果我试试 sb.append((lseq.charAt(i) ^ tseq.charAt(i) ^ hseq.charAt(i));

我得到了结果:

48484848484848484848484948484948494848494848494949484848494848494848494849494848484949494948484848484948494948494949484948484948

这是错的.我需要帮助做3个二进制字符串的xor

Answer 1

我会这样做的

private static boolean bitOf(char in) {
    return (in == '1');
}

private static char charOf(boolean in) {
    return (in) ? '1' : '0';
}

public static void main(String[] args) {
    String lseq ="0011111111101111111111111100101101111100110000001011111000010100";
    String tseq ="0011111111100000110011001100110011001100110011001100110011001100";
    String hseq ="0011111111110010010111110100010111100000101101001110000100011110";

    StringBuilder sb = new StringBuilder();

    for (int i = 0; i < lseq.length(); i++) {
        sb.append(charOf(bitOf(lseq.charAt(i)) ^ bitOf(tseq.charAt(i))
                ^ bitOf(hseq.charAt(i))));
    }

    String result = sb.toString();
    System.out.println(result);
}

哪个输出

0011111111111101011011000100001001010000101110001001001111000110