3

我有以下函数来计算字符串中逗号（或任何其他字符）的数量，而不计算双引号内的逗号。我想知道是否有更好的方法来实现这一点，或者即使你能找到这个函数可能崩溃的情况。

public int countCharOfString(char c, String s) {
    int numberOfC = 0;
    boolean doubleQuotesFound = false;
    for(int i = 0; i < s.length(); i++){
        if(s.charAt(i) == c && !doubleQuotesFound){
            numberOfC++;
        }else if(s.charAt(i) == c && doubleQuotesFound){
            continue;
        }else if(s.charAt(i) == '\"'){
            doubleQuotesFound = !doubleQuotesFound;
        }
    }
    return numberOfC;
}

感谢您的任何建议

4

7 回答 7

4

此实现有两个不同之处：

  • 使用CharSequence而不是字符串
  • boolean如果我们在引用的子序列中，则不需要跟踪值。

功能：

public static int countCharOfString(char quote, CharSequence sequence) {

    int total = 0, length = sequence.length();

    for(int i = 0; i < length; i++){
        char c = sequence.charAt(i);
        if (c == '"') {
            // Skip quoted sequence
            for (i++; i < length && sequence.charAt(i)!='"'; i++) {}
        } else if (c == quote) {
            total++;
        }
    }

    return total;
 }
于 2012-04-11T20:04:50.043 回答
2 
public static int countCharOfString(char c, String s)
{
    int numberOfC = 0;
    int innerC = 0;
    boolean holdDoubleQuotes = false;
    for(int i = 0; i < s.length(); i++)
    {
        char r = s.charAt(i);
        if(i == s.length() - 1 && r != '\"')
        {
            numberOfC += innerC;
            if(r == c) numberOfC++;
        }
        else if(r == c && !holdDoubleQuotes) numberOfC++;
        else if(r == c && holdDoubleQuotes) innerC++;
        else if(r == '\"' && holdDoubleQuotes)
        {
            holdDoubleQuotes = false;
            innerC = 0;
        }
        else if(r == '\"' && !holdDoubleQuotes) holdDoubleQuotes = true;
    }
    return numberOfC;
}


System.out.println(countCharOfString(',', "Hello, BRabbit27, how\",,,\" are, you?"));

输出：

3

另一种方法是使用正则表达式：

public static int countCharOfString(char c, String s)
{
   s = " " + s + " "; // To make the first and last commas to be counted
   return s.split("[^\"" + c + "*\"][" + c + "]").length - 1;
}
于 2012-04-11T19:38:03.220 回答
1
  • 你不应该charAt()在循环内多次调用。使用char变量。
  • 您不应该length()为每次迭代调用。int在循环之前使用。
  • 你应该避免重复比较c- 使用嵌套的 if/else。
于 2012-04-11T19:39:02.063 回答
1

需要很大的字符串才能产生很大的不同。

这段代码更快的原因是它平均每个循环包含 1.5 个检查，而不是每个循环 3 个检查。它通过使用两个循环来做到这一点，一个用于引用状态，一个用于未引用状态。

public static void main(String... args) {
    String s = generateString(20 * 1024 * 1024);
    for (int i = 0; i < 15; i++) {
        long start = System.nanoTime();
        countCharOfString(',', s);
        long mid = System.nanoTime();
        countCharOfString2(',', s);
        long end = System.nanoTime();
        System.out.printf("countCharOfString() took %.3f ms, countCharOfString2() took %.3f ms%n",
                (mid - start) / 1e6, (end - mid) / 1e6);
    }
}

private static String generateString(int length) {
    StringBuilder sb = new StringBuilder(length);
    Random rand = new Random(1);
    while (sb.length() < length)
        sb.append((char) (rand.nextInt(96) + 32)); // includes , and "
    return sb.toString();
}

public static int countCharOfString2(char c, String s) {
    int numberOfC = 0, i = 0;
    while (i < s.length()) {
        // not quoted
        while (i < s.length()) {
            char ch = s.charAt(i++);
            if (ch == c)
                numberOfC++;
            else if (ch == '"')
                break;
        }
        // quoted
        while (i < s.length()) {
            char ch = s.charAt(i++);
            if (ch == '"')
                break;
        }
    }
    return numberOfC;
}


public static int countCharOfString(char c, String s) {
    int numberOfC = 0;
    boolean doubleQuotesFound = false;
    for (int i = 0; i < s.length(); i++) {
        if (s.charAt(i) == c && !doubleQuotesFound) {
            numberOfC++;
        } else if (s.charAt(i) == c && doubleQuotesFound) {
            continue;
        } else if (s.charAt(i) == '\"') {
            doubleQuotesFound = !doubleQuotesFound;
        }
    }
    return numberOfC;
}

印刷

countCharOfString() took 33.348 ms, countCharOfString2() took 31.381 ms
countCharOfString() took 28.265 ms, countCharOfString2() took 25.801 ms
countCharOfString() took 28.142 ms, countCharOfString2() took 14.576 ms
countCharOfString() took 28.372 ms, countCharOfString2() took 14.540 ms
countCharOfString() took 28.191 ms, countCharOfString2() took 14.616 ms
于 2012-04-11T20:07:46.883 回答
1

也许不是最快的...

public int countCharOfString(char c, String s) {
    final String removedQuoted = s.replaceAll("\".*?\"", "");
    int total = 0;
    for(int i = 0; i < removedQuoted.length(); ++i)
        if(removedQuoted.charAt(i) == c)
            ++total;
    return total;
}
于 2012-04-11T19:48:56.843 回答
1

更简单，更不容易出错（是的，性能低于逐个字符遍历字符串并手动跟踪所有内容）：

public static int countCharOfString(char c, String s) {
  s = s.replaceAll("\".*?\"", "");
  int cnt = 0;
  for (int foundAt = s.indexOf(c); foundAt > -1; foundAt = s.indexOf(c, foundAt+1)) 
    cnt++;
  return cnt;
}
于 2012-04-11T20:12:08.653 回答
0

您还可以使用正则表达式和 String.split()

它可能看起来像这样：

public int countNonQuotedOccurrences(String inputstring, char searchChar)
{
  String regexPattern = "[^\"]" + searchChar + "[^\"]";
  return inputString.split(regexPattern).length - 1;
}

免责声明：

这只是显示了基本方法。

上面的代码不会在字符串的开头或结尾检查 searchChar。

您可以手动检查或添加到 regexPattern。

于 2012-04-11T19:48:06.907 回答