3

如何编写一个程序,它将以001010101110000100100...., 011100010001000011000...., 000000000010000000000100.... 作为输入(位),输出将是OR这 3 个。

OR = 0 0 = 0,
     0 1 = 1,
     1 0 = 1,
     1 1 = 1,

如果 sombody 有一个也会有帮助的示例程序。我们需要将值从字节存储在位数组中吗?

4

2 回答 2

4

你不能只调用orBitSet 类中的方法吗?

[编辑]假设你想要一个例子,这样的事情应该可以工作:

BitSet doOr( List<BitSet> setsToOr ) {
  BitSet ret = null ;
  for( BitSet set : setsToOr ) {
    if( ret == null ) {
      // Set ret to a copy of the first set in the list
      ret = (BitSet)set.clone() ;
    }
    else {
      // Just or with the current set (changes the value of ret)
      ret.or( set ) ;
    }
  }
  // return the result
  return ret ;
}
于 2011-03-09T11:01:42.227 回答
1

这应该有效(更新:已修复错误):

public static BitSet or(final String... args){
    final BitSet temp = createBitset(args[0]);
    for(int i = 1; i < args.length; i++){
        temp.or(createBitset(args[i]));
    }
    return temp;
}

private static BitSet createBitset(final String input){
    int length = input.length();
    final BitSet bitSet = new BitSet(length);
    for(int i = 0; i < length; i++){
        // anything that's not a 1 is a zero, per convention
        bitSet.set(i, input.charAt(length - (i + 1)) == '1');
    }
    return bitSet;
}

示例代码:

public static void main(final String[] args){
    final BitSet bs =
        or("01010101", "10100000", "00001010", "1000000000000000");
    System.out.println(bs);
    System.out.println(toCharArray(bs));
}

private static char[] toCharArray(final BitSet bs){
    final int length = bs.length();
    final char[] arr = new char[length];
    for(int i = 0; i < length; i++){
        arr[i] = bs.get(i) ? '1' : '0';
    }
    return arr;
}

输出:

{0、1、2、3、4、5、6、7、15}
1111111100000001

于 2011-03-09T11:12:07.903 回答