我将首先解释我的意思是“补充不包括前导零二进制位的整数值”(从现在开始,为简洁起见,我将其称为非前导零位补码或NLZ补码)。
例如,有整数92.二进制数是1011100.如果我们执行正常的按位NOT或补码,结果是:-93(有符号整数)或11111111111111111111111110100011(二进制)。那是因为前导零位也在补充。
因此,对于NLZ补码,前导零位不补充,那么NLZ补充92或1011100的结果是:35或100011(二进制)。通过将输入值与非前导零值的1位序列进行异或来执行操作。插图:
92: 1011100
1111111 (xor)
--------
0100011 => 35
我做了这样的java算法:
public static int nonLeadingZeroComplement(int n) {
if (n == 0) {
return ~n;
}
if (n == 1) {
return 0;
}
//This line is to find how much the non-leading zero (NLZ) bits count.
//This operation is same like: ceil(log2(n))
int binaryBitsCount = Integer.SIZE - Integer.numberOfLeadingZeros(n - 1);
//We use the NLZ bits count to generate sequence of 1 bits as much as the NLZ bits count as complementer
//by using shift left trick that equivalent to: 2 raised to power of binaryBitsCount.
//1L is one value with Long literal that used here because there is possibility binaryBitsCount is 32
//(if the input is -1 for example), thus it will produce 2^32 result whom value can't be contained in
//java signed int type.
int oneBitsSequence = (int)((1L << binaryBitsCount) - 1);
//XORing the input value with the sequence of 1 bits
return n ^ oneBitsSequence;
}
我需要一个如何优化上述算法的建议,特别是生成1位补充序列(oneBitsSequence)的行,或者是否有人能提出更好的算法?
更新:我也想知道这个非领先的零补充的已知术语?
答案 0 :(得分:8)
您可以通过Integer.highestOneBit(i)
方法获得最高的一位,向左移一步,然后减去1.这样可以得到1
的正确长度:
private static int nonLeadingZeroComplement(int i) {
int ones = (Integer.highestOneBit(i) << 1) - 1;
return i ^ ones;
}
例如,
System.out.println(nonLeadingZeroComplement(92));
打印
35
答案 1 :(得分:2)
显然@keppil提供了最短的解决方案。另一种解决方案可能就像。
private static int integerComplement(int n){
String binaryString = Integer.toBinaryString(n);
String temp = "";
for(char c: binaryString.toCharArray()){
if(c == '1'){
temp += "0";
}
else{
temp += "1";
}
}
int base = 2;
int complement = Integer.parseInt(temp, base);
return complement;
}
例如,
System.out.println(nonLeadingZeroComplement(92));
将答案打印为35