java求教2个任意相同长度的（01）字符串如何快速的进行'与'或者'或'操作
例如 0101 | 1110 = 1111 （要任意长度字符串 效率第一的）
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- Java code
public static String and(String str1, String str2) { StringBuffer sb = new StringBuffer(str1.length()); for (int i = 0; i < str1.length(); i++) { if (str1.charAt(i) == str2.charAt(i) && str2.charAt(i) == '1') { sb.append("1"); } else { sb.append("0"); } } return sb.toString(); } public static String or(String str1, String str2) { StringBuffer sb = new StringBuffer(str1.length()); for (int i = 0; i < str1.length(); i++) { if (str1.charAt(i) == '1' || str2.charAt(i) == '1') { sb.append("1"); } else { sb.append("0"); } } return sb.toString(); }
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for example
- Java code
System.out.println(and("0101", "1110")); System.out.println(or("0101", "1110")); System.out.println(xor("0101", "1110")); System.out.println(not("0101")); public static String and(String s1, String s2) { return new BigInteger(s1, 2).and(new BigInteger(s2, 2)).toString(2); } public static String or(String s1, String s2) { return new BigInteger(s1, 2).or(new BigInteger(s2, 2)).toString(2); } public static String xor(String s1, String s2) { return new BigInteger(s1, 2).xor(new BigInteger(s2, 2)).toString(2); } public static String not(String s) { return new BigInteger(s, 2).not().toString(2); }