Question

我刚刚完成了一个关于Kattis的问题＆＃34;一个理性序列2＆＃34;并想知道是否有更有效的方法可以将二进制数转换为十进制数。这是我的代码：

public static int getDecimalValue(String sequence){

    int value = 1;

    for(int i = 0; i < sequence.length(); i++){
        if(sequence.charAt(i) == '1')
            value += (int)Math.pow(2, i);
    }
    return value;
}

任何帮助都会很棒！

Answer 1

int value = Integer.parseInt（sequence，2）;

Answer 2

有几点。首先，你的代码实际上是一个错误 - 尝试传递它00000000（8个零）并看看会发生什么。

至于效率，你可以节省几笔钱。你可以改变你计算长度的位置，你可以bitshift计算，这要快得多。

public static int getBinaryValue(String sequence){

    int value = 1; //have another glance at this line!

    for(int i = 0, n=sequence.length(); i < n; i++){
    //I declared a variable 'n' in the initialisation, this means its only 
    //checked once, rather than being checked every time
        if(sequence.charAt(i) == '1')
            value += 1 << i;
            //and here I've bitshifted the value.  Basically I've said "take 
            //the number one and then shift it left down an imaginary binary
            //track i times".  So if i is three, for example, it'll shift it
            //from 00000001 to 00000010 to 00000100 to 00001000, which is 8
            //2^3 = 8
    }
    return value;
}

更高效的解决方案

2 个答案: