【发布时间】:2018-11-05 15:48:06
【问题描述】:
我成功完成了以下任务,但是代码效率非常低,如果有人能告诉我一个更有效的方法,我将不胜感激,也许使用子字符串。作业:
/**
* Separates a given string into tokens, which are the "words" that are
* separated by one or more occurrences of the given separator character.
* Returns the tokens as an array of String values.
*/
public static String[] tokenize (String str, char separator) {
// Removes all the occurrences of the separator at the beginning and end of str
String source = trim(str, separator);
String[] tokens = new String[charRunCount (source,separator)+1];
String tmp = ""; // a string in order to take a word, then run over this string
int j = 0;
int i = 0;
while (i < tokens.length) {
if ( source.charAt (j) != separator ) {
do {
tmp += source.charAt (j);
if ( j >= source.length () - 1 ) {
break;
}
else { // so that we math the source length
j++;
}
} while (source.charAt (j) != separator);
}
if ( source.charAt (j) == separator ) {
j++;
while (source.charAt (j) == separator) {
j++;
}
}
tokens[i] = tmp;// taking the token into place
tmp = ""; //resetting the token so we can begin anew
i++;
}
return tokens;
}
cahrRunCount 函数:
public static int charRunCount(String str, char c){
char last = 0;
int counter = 0;
for (int i = 0; i < str.length(); i++) {
// whenever a run starts.
if (last != c && str.charAt(i) == c) {
counter++;
}
last = str.charAt(i);
}
return counter;
}
我不能使用 import 或 regex,谢谢!
【问题讨论】:
-
如果您的代码有效但需要改进,CodeReview 可能是一个更好的发布位置。
-
对不起,我不知道有这样的网站存在