Java中二进制字符串的Xor [英] Xor of binary strings in java
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问题描述
我使用以下代码来XOR 2字符串
i am using the following code to XOR 2 strings
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:
The above code gives me correct output :
0000000000001111001100110000011110110000000011000111001011011000
我需要多扫一遍字符串
String hseq = "0011111111110010010111110100010111100000101101001110000100011110";
如果我尝试 sb.append((lseq.charAt(i)^ tseq.charAt(i)^ hseq.charAt(i));
我得到结果:
48484848484848484848484948484948494848494848494949484848494848494848494849494848484949494948484848484948494948494949484948484948
这是错误的。我需要帮助做3个二进制字符串的xor
which is wrong. I need help in doing xor of 3 binary strings
推荐答案
我会这样做
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);
}
哪些输出
0011111111111101011011000100001001010000101110001001001111000110
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