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？

  1. static String or_str(String s1, String s2){
  2.     StringBuilder result = new StringBuilder();
  3.     int length = Math.min(s1.length(), s2.length()); // 取两个字符串长度的较小值
  4.     
  5.     for(int i = 0; i < length; i++){
  6.         if(s1.charAt(i) == '1' || s2.charAt(i) == '1') {
  7.             result.append("1");
  8.         } else {
  9.             result.append("0"); // 如果两个字符都不是 '1'，则按位或结果为 '0'
  10.         }
  11.     }
  12.     
  13.     return result.toString();
  14. }
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?

  1.  static String or_str(String s1, String s2){
  2.             String result=&quot;&quot;;
  3.             for(int i=0; i&lt;s1.length(); i++){
  4.                 if(s1.charAt(i)==&#39;1&#39;||s2.charAt(i)==&#39;1&#39;)
  5.                 result+=&quot;1&quot;;
  6.             }
  7.             return result;
  8.         }

答案1

得分: 2

  1. 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：
  3. static int countBitwiseOr(String s1, String s2) {
  4.     int result = 0;
  5.     for (int i = 0; i < s1.length(); i++) {
  6.         if (s1.charAt(i) == '1' || s2.charAt(i) == '1')
  7.             result += 1;
  8.     }
  9.     return result;
  10. }
  12. 以避免大量的计算开销。
  14. **在评论中提出的建议的符合API的编辑解决方案**
  16. 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：
  18. > 使用Java 11
  20. static String or_str(String s1, String s2){
  21.     return "1".repeat(countBitwiseOr(s1, s2));
  22. }
  24. > 使用Java 11
  26. static String or_str(String s1, String s2){
  27.     return Stream.generate(() -> "1").limit(countBitwiseOr(s1, s2)).collect(joining());
  28. }
英文:

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

  1. static int countBitwiseOr(String s1, String s2) {
  2. 	int result = 0;
  3. 	for (int i = 0; i &lt; s1.length(); i++) {
  4. 		if (s1.charAt(i) == &#39;1&#39; || s2.charAt(i) == &#39;1&#39;)
  5. 			result += 1;
  6. 	}
  7. 	return result;
  8. }

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

  1.  static String or_str(String s1, String s2){
  2.     return &quot;1&quot;.repeat(countBitwiseOr(s1,s2));
  3.  }

> with Java 11

  1.  static String or_str(String s1, String s2){
  2.    return Stream.generate(() -&gt; &quot;1&quot;).limit(countBitwiseOr(s1,s2)).collect(joining());
  3.  }

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：

  1. public static void main(String[] args) {
  2.     String s1 = "1011010100100101110";
  3.     String s2 = "011011101101001000100110001";
  4.     System.out.println(or_str(s1, s2));
  5. }
  7. static String or_str(String s1, String s2){
  8.     char[] res = new char[s1.length()];
  9.     for(int i=0; i<s1.length(); i++){
  10.         if(s1.charAt(i)=='1'||s2.charAt(i)=='1')
  11.             res[i] ='1';
  12.         else
  13.             res[i] ='0';
  14.     }
  15.     return new String(res);
  16. }

如果您运行这个代码，输出将会是：

  1. 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:

  1. public static void main(String[] args) {
  2.     String s1 = &quot;1011010100100101110&quot;;
  3.     String s2 = &quot;011011101101001000100110001&quot;;
  4.     System.out.println(or_str(s1, s2));
  5. }
  7. static String or_str(String s1, String s2){
  8.     char[] res = new char[s1.length()];
  9.     for(int i=0; i&lt;s1.length(); i++){
  10.         if(s1.charAt(i)==&#39;1&#39;||s2.charAt(i)==&#39;1&#39;)
  11.             res[i] =&#39;1&#39;;
  12.         else
  13.             res[i] =&#39;0&#39;;
  14.     }
  15.     return new String(res);
  16. }

If you run this, the output will be:

  1. 1111111111110111111

huangapple
  • 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
  • bitwise-or
  • java
  • string
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