Java中二进制字符串的Xor
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";
如果我尝试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
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
您可以使用BigInteger ,它将使您的代码更简单。 它有一个构造函数,您可以在其中传递String和要使用的基础(在您的情况下为2)。
然后你可以做一个XOR调用public BigInteger xor(BigInteger val)
(参见docs或者看看openjdk BigInteger的代码 )