How to optimize the following code in java where I've to get bitwise OR of two very long binary strings?

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英文:

How to optimize the following code in java where I've to get bitwise OR of two very long binary strings?

问题

我已经写了以下代码来获取两个字符串的**按位或(bitwise or)**操作结果。但是,当我传递较长的字符串时,执行时间太长。我该如何进行优化

static String or_str(String s1, String s2){
    StringBuilder result = new StringBuilder();
    int length = Math.min(s1.length(), s2.length()); // 取两个字符串长度的较小值
    
    for(int i = 0; i < length; i++){
        if(s1.charAt(i) == '1' || s2.charAt(i) == '1') {
            result.append("1");
        } else {
            result.append("0"); // 如果两个字符都不是 '1',则按位或结果为 '0'
        }
    }
    
    return result.toString();
}
英文:

I've written the following code to get the bitwise or of two strings. But, it's taking way too long when I pass long strings. How can I optimize it?

 static String or_str(String s1, String s2){
            String result=&quot;&quot;;
            for(int i=0; i&lt;s1.length(); i++){
                if(s1.charAt(i)==&#39;1&#39;||s2.charAt(i)==&#39;1&#39;)
                result+=&quot;1&quot;;
            }
            return result;
        }

答案1

得分: 2

如果您只需要计算1的数量就像评论中提到的那样您可以使用以下示例代码

static int countBitwiseOr(String s1, String s2) {
    int result = 0;
    for (int i = 0; i < s1.length(); i++) {
        if (s1.charAt(i) == '1' || s2.charAt(i) == '1')
            result += 1;
    }
    return result;
}

以避免大量的计算开销

**在评论中提出的建议的符合API的编辑解决方案**

如果您需要符合API的结果返回包含所有正结果的字符串),您仍然可以

> 使用Java 11

static String or_str(String s1, String s2){
    return "1".repeat(countBitwiseOr(s1, s2));
}

> 使用Java 11

static String or_str(String s1, String s2){
    return Stream.generate(() -> "1").limit(countBitwiseOr(s1, s2)).collect(joining());
}
英文:

If you need just the amount of ones, like can be found on comments, you can use sample like this:

static int countBitwiseOr(String s1, String s2) {
	int result = 0;
	for (int i = 0; i &lt; s1.length(); i++) {
		if (s1.charAt(i) == &#39;1&#39; || s2.charAt(i) == &#39;1&#39;)
			result += 1;
	}
	return result;
}

to avoid lot of calculation overhead.

Edit suggested API conform solution proposed in comments

If you need to rest API conform (return a string containing all the positive results) you then still can:

> with Java 11

 static String or_str(String s1, String s2){
    return &quot;1&quot;.repeat(countBitwiseOr(s1,s2));
 }

> with Java 11

 static String or_str(String s1, String s2){
   return Stream.generate(() -&gt; &quot;1&quot;).limit(countBitwiseOr(s1,s2)).collect(joining());
 }

答案2

得分: 0

如果您想要一个真正的位“或”操作,您还需要在条件不匹配的情况下输出零。如果您想提高性能,您可以知道输出的长度,因此在每次迭代时,可以从开始就使用一个预定义长度的字符数组。

以下是一个简单的字符串位或实现,它返回一个字符串:

public static void main(String[] args) {
    String s1 = "1011010100100101110";
    String s2 = "011011101101001000100110001";
    System.out.println(or_str(s1, s2));
}

static String or_str(String s1, String s2){
    char[] res = new char[s1.length()];
    for(int i=0; i<s1.length(); i++){
        if(s1.charAt(i)=='1'||s2.charAt(i)=='1')
            res[i] ='1';
        else
            res[i] ='0';
    }
    return new String(res);
}

如果您运行这个代码,输出将会是:

1111111111110111111
英文:

If you want a real bitwise "OR" you need also to output a zero in case your condition does not match. If you want to improve performance you know the length of the output so instead of creating a new string on each iteration you can use a character array with a pre-defined length from the start.

Here is a simple implementation of a bitwise or for strings which returns a string:

public static void main(String[] args) {
    String s1 = &quot;1011010100100101110&quot;;
    String s2 = &quot;011011101101001000100110001&quot;;
    System.out.println(or_str(s1, s2));
}

static String or_str(String s1, String s2){
    char[] res = new char[s1.length()];
    for(int i=0; i&lt;s1.length(); i++){
        if(s1.charAt(i)==&#39;1&#39;||s2.charAt(i)==&#39;1&#39;)
            res[i] =&#39;1&#39;;
        else
            res[i] =&#39;0&#39;;
    }
    return new String(res);
}

If you run this, the output will be:

1111111111110111111

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  • 本文由 发表于 2020年3月16日 20:06:33
  • 转载请务必保留本文链接:https://go.coder-hub.com/60705695.html
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