[英]Would there be an approach without using a regular expression?
This is an interview question from http://www.glassdoor.com/Interview/Indeed-Software-Engineer-Interview-Questions-EI_IE100561.0,6_KO7,24.htm , specifically, the problem 这是来自http://www.glassdoor.com/Interview/Indeed-Software-Engineer-Interview-Questions-EI_IE100561.0,6_KO7,24.htm的采访问题,具体来说就是问题
"The asked me a method that took in a string and return words up to the max length. There could be any amount of spaces which made it a little tricky " “问我一个方法,该方法接受一个字符串并返回最大长度的单词。可能有任意数量的空格,这使它有点棘手”
Here is my solution(with test case) 这是我的解决方案(带有测试用例)
public class MaxLength {
public static void main(String[] args) {
List<String> allWords = maxWords("Jasmine has no love for chris", 2);
for(String word: allWords){
System.out.println(word);
}
}
public static List<String> maxWords(String sentence, int length) {
String[] words = sentence.trim().split("\\s+");
List<String> list = new ArrayList<String>();
for(String word: words) {
if(word.length() <= length) {
list.add(word);
}
}
return list;
}
The test ran fine and i got my expected output - no. 测试进行得很好,我得到了预期的输出-不。 However during an actual interview, I think the interviewer doesn't expect you to know this regular expression off the top of your head (I didn't had to find it from How do I split a string with any whitespace chars as delimiters? ) Is there another approach to this problem without using regular expressions? 但是在实际面试中,我认为面试官并不希望您知道这个正则表达式(您不必从我如何用任何空白字符作为分隔符的字符串拆分中找到它)。在不使用正则表达式的情况下,还有另一种方法可以解决此问题吗?
You could use the StringTokenizer
. 您可以使用StringTokenizer
。
Another, maybe more lengthy approach would be to iterate over the string and use the methods provided by the Character class ( Character.isWhiteSpace(char c)
) and break the string accordingly. 另一种可能更长的方法是遍历字符串并使用Character类( Character.isWhiteSpace(char c)
)提供的方法并相应地中断字符串。
Try: 尝试:
public static List<String> maxWords(String sentence, int length) {
List<String> list = new ArrayList<String>();
String tmp = sentence;
while (tmp.indexOf(" ") != -1) {
String word = tmp.substring(0,tmp.indexOf(" "));
tmp = tmp.substring(tmp.indexOf(" ")+1);
if(word.length() <= length) {
list.add(word);
}
}
return list;
}
Well, you could try something like this : Doesn't create many subStrings()
just creates subStrings of "matched" Strings 好吧,您可以尝试执行以下操作:不创建许多subStrings()
而是创建“匹配”字符串的子字符串
public class Test {
public static void main(String[] args) {
String s = "Jasmine has no love for chris";
s=s.trim();
List<String> ls = getMaxLength(s, 3);
for (String str : ls) {
System.out.println(str);
}
}
static List<String> getMaxLength(String s, int length) {
List<String> ls = new ArrayList<String>();
int i = 0;
if (s.charAt(i + length) == ' ' || i + length == s.length()) { // search if first word matches your criterion.
for (i = 1; i < length; i++) { // check for whitespace between 0 and length
if (s.charAt(i) == ' ') {
i = length;
break;
}
if (i == length - 1)
ls.add(s.substring(0, length));
}
}
for (i = length; i < s.length(); i++) {// start search from second word..
if (s.charAt(i - 1) == ' ' && i + length < s.length() // if char at length is space or end of String, make sure the char before the current char is a space.
&& (s.charAt(i + length) == ' ' || i + length == s.length())) {
for (int j = i; j < i + length; j++) {
if (s.charAt(j) == ' ') {
i = i + length;
break;
}
if (j == i + length - 1) {
ls.add(s.substring(i, i + length));
}
}
}
}
return ls;
} // end of method
}
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