Question

如何编写一个程序，将001010101110000100100 ....，011100010001000011000 ....，000000000010000000000100 ....作为输入（位），输出将是这些中的OR

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

如果sombody有一个样本程序也会有帮助。我们是否需要将值存储在来自byte？

的位数组中

Answer 1

你不能只拨打or method in the BitSet class吗？

[edit]假设你想要一个例子，这样的事情应该有效：

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 ;
}

Answer 2

这应该可行（更新：错误已修复）：

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;
}

<强>输出：

{0,1,2,3,4,5,6,7,15}
1111111100000001

Java中的OR操作（BitSet.class）

2 个答案: