假设我有两个字符串，
String s1 = "AbBaCca";
String s2 = "bac";
我想执行一个检查，返回s2包含在s1中。我可以这样做:
return s1.contains(s2);
我非常确定contains()是区分大小写的，但是我不能从阅读文档中确定这一点。如果是的话，我想我最好的方法是:
return s1.toLowerCase().contains(s2.toLowerCase());
撇开所有这些不谈，有没有另一种(可能更好的)方法可以在不考虑大小写敏感性的情况下完成这个任务?
当前回答
如果你必须在另一个ASCII字符串中搜索一个ASCII字符串，比如URL，你会发现我的解决方案更好。我测试了icza和我的方法的速度，结果如下:
案例1用了2788 ms - regionMatches 病例2用了1520毫秒
代码:
public static String lowerCaseAscii(String s) {
if (s == null)
return null;
int len = s.length();
char[] buf = new char[len];
s.getChars(0, len, buf, 0);
for (int i=0; i<len; i++) {
if (buf[i] >= 'A' && buf[i] <= 'Z')
buf[i] += 0x20;
}
return new String(buf);
}
public static boolean containsIgnoreCaseAscii(String str, String searchStr) {
return StringUtils.contains(lowerCaseAscii(str), lowerCaseAscii(searchStr));
}
其他回答
你可以使用正则表达式，它工作:
boolean found = s1.matches("(?i).*" + s2+ ".*");
更快的实现:使用String.regionMatches()
使用regexp可能相对较慢。如果你只是想检查一种情况，(慢)没有关系。但如果你有一个数组或一个包含成千上万个字符串的集合，事情就会变得非常缓慢。
下面给出的解决方案既不使用正则表达式，也不使用toLowerCase()(这也很慢，因为它创建了另一个字符串，并在检查后丢弃它们)。
解决方案构建在String.regionMatches()方法上，该方法似乎是未知的。它检查2个String区域是否匹配，但重要的是它还有一个重载，带有一个方便的ignoreCase参数。
public static boolean containsIgnoreCase(String src, String what) {
final int length = what.length();
if (length == 0)
return true; // Empty string is contained
final char firstLo = Character.toLowerCase(what.charAt(0));
final char firstUp = Character.toUpperCase(what.charAt(0));
for (int i = src.length() - length; i >= 0; i--) {
// Quick check before calling the more expensive regionMatches() method:
final char ch = src.charAt(i);
if (ch != firstLo && ch != firstUp)
continue;
if (src.regionMatches(true, i, what, 0, length))
return true;
}
return false;
}
速度分析
这种速度分析并不意味着是火箭科学，只是对不同方法的速度有多快的粗略描述。
我比较了5种方法。
Our containsIgnoreCase() method. By converting both strings to lower-case and call String.contains(). By converting source string to lower-case and call String.contains() with the pre-cached, lower-cased substring. This solution is already not as flexible because it tests a predefiend substring. Using regular expression (the accepted answer Pattern.compile().matcher().find()...) Using regular expression but with pre-created and cached Pattern. This solution is already not as flexible because it tests a predefined substring.
结果(通过调用该方法1000万次):
我们的方法是670毫秒 2x toLowerCase() and contains(): 2829 ms 1x toLowerCase()和contains()，缓存子字符串:2446毫秒 Regexp: 7180 ms 缓存模式的Regexp: 1845毫秒
表格中的结果:
RELATIVE SPEED 1/RELATIVE SPEED
METHOD EXEC TIME TO SLOWEST TO FASTEST (#1)
------------------------------------------------------------------------------
1. Using regionMatches() 670 ms 10.7x 1.0x
2. 2x lowercase+contains 2829 ms 2.5x 4.2x
3. 1x lowercase+contains cache 2446 ms 2.9x 3.7x
4. Regexp 7180 ms 1.0x 10.7x
5. Regexp+cached pattern 1845 ms 3.9x 2.8x
我们的方法比使用小写和contains()快4倍，比使用正则表达式快10倍，即使Pattern是预缓存的也快3倍(并且失去了检查任意子字符串的灵活性)。
分析测试代码
如果你对分析是如何执行的感兴趣，下面是完整的可运行应用程序:
import java.util.regex.Pattern;
public class ContainsAnalysis {
// Case 1 utilizing String.regionMatches()
public static boolean containsIgnoreCase(String src, String what) {
final int length = what.length();
if (length == 0)
return true; // Empty string is contained
final char firstLo = Character.toLowerCase(what.charAt(0));
final char firstUp = Character.toUpperCase(what.charAt(0));
for (int i = src.length() - length; i >= 0; i--) {
// Quick check before calling the more expensive regionMatches()
// method:
final char ch = src.charAt(i);
if (ch != firstLo && ch != firstUp)
continue;
if (src.regionMatches(true, i, what, 0, length))
return true;
}
return false;
}
// Case 2 with 2x toLowerCase() and contains()
public static boolean containsConverting(String src, String what) {
return src.toLowerCase().contains(what.toLowerCase());
}
// The cached substring for case 3
private static final String S = "i am".toLowerCase();
// Case 3 with pre-cached substring and 1x toLowerCase() and contains()
public static boolean containsConverting(String src) {
return src.toLowerCase().contains(S);
}
// Case 4 with regexp
public static boolean containsIgnoreCaseRegexp(String src, String what) {
return Pattern.compile(Pattern.quote(what), Pattern.CASE_INSENSITIVE)
.matcher(src).find();
}
// The cached pattern for case 5
private static final Pattern P = Pattern.compile(
Pattern.quote("i am"), Pattern.CASE_INSENSITIVE);
// Case 5 with pre-cached Pattern
public static boolean containsIgnoreCaseRegexp(String src) {
return P.matcher(src).find();
}
// Main method: perfroms speed analysis on different contains methods
// (case ignored)
public static void main(String[] args) throws Exception {
final String src = "Hi, I am Adam";
final String what = "i am";
long start, end;
final int N = 10_000_000;
start = System.nanoTime();
for (int i = 0; i < N; i++)
containsIgnoreCase(src, what);
end = System.nanoTime();
System.out.println("Case 1 took " + ((end - start) / 1000000) + "ms");
start = System.nanoTime();
for (int i = 0; i < N; i++)
containsConverting(src, what);
end = System.nanoTime();
System.out.println("Case 2 took " + ((end - start) / 1000000) + "ms");
start = System.nanoTime();
for (int i = 0; i < N; i++)
containsConverting(src);
end = System.nanoTime();
System.out.println("Case 3 took " + ((end - start) / 1000000) + "ms");
start = System.nanoTime();
for (int i = 0; i < N; i++)
containsIgnoreCaseRegexp(src, what);
end = System.nanoTime();
System.out.println("Case 4 took " + ((end - start) / 1000000) + "ms");
start = System.nanoTime();
for (int i = 0; i < N; i++)
containsIgnoreCaseRegexp(src);
end = System.nanoTime();
System.out.println("Case 5 took " + ((end - start) / 1000000) + "ms");
}
}
有一个简单简洁的方法，使用regex标志(不区分大小写{i}):
String s1 = "hello abc efg";
String s2 = "ABC";
s1.matches(".*(?i)"+s2+".*");
/*
* .* denotes every character except line break
* (?i) denotes case insensitivity flag enabled for s2 (String)
* */
我做了一个测试，找到一个字符串的大小写不敏感匹配。我有一个150000个对象的向量，所有对象都有一个字符串作为一个字段，并想找到匹配字符串的子集。我尝试了三种方法:
Convert all to lower case for (SongInformation song: songs) { if (song.artist.toLowerCase().indexOf(pattern.toLowercase() > -1) { ... } } Use the String matches() method for (SongInformation song: songs) { if (song.artist.matches("(?i).*" + pattern + ".*")) { ... } } Use regular expressions Pattern p = Pattern.compile(pattern, Pattern.CASE_INSENSITIVE); Matcher m = p.matcher(""); for (SongInformation song: songs) { m.reset(song.artist); if (m.find()) { ... } }
定时结果为:
没有尝试匹配:20毫秒 低匹配:182毫秒 字符串匹配:278毫秒 正则表达式:65毫秒
对于这个用例，正则表达式看起来是最快的。
我不确定你的主要问题是什么，但是是的，.contains是区分大小写的。
