标题翻译
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();
}
英文翻译
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="";
for(int i=0; i<s1.length(); i++){
if(s1.charAt(i)=='1'||s2.charAt(i)=='1')
result+="1";
}
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());
}
英文翻译
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 < s1.length(); i++) {
if (s1.charAt(i) == '1' || s2.charAt(i) == '1')
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 "1".repeat(countBitwiseOr(s1,s2));
}
> with Java 11
static String or_str(String s1, String s2){
return Stream.generate(() -> "1").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
英文翻译
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 = "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);
}
If you run this, the output will be:
1111111111110111111
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