返回字符串“code”在给定字符串中任何位置出现的次数

Question

public int countCode(String str) {
  int code = 0;
  
  for(int i=0; i<str.length()-3; i++){
    if(str.substring(i, i+2).equals("co") && str.charAt(i+3)=='e'){
      code++;
    }
  }
  return code;
}

Hi guys, I've solved this problem by some help among the inte.net. But the actual problem that I'm facing is this, (str.length()-3) in the for loop. I don't understand why the str.length()-3 having this -3 in it. please explain it...

Answer 1

Inside the for loop, for any index ( i ), it checks that the chars at i and i+2 and i+3 match your requirements. If your i becomes length of your string (or last character), then the code will throw exception since it will try to find char at position which is not really there.

Answer 2

In the interest of posting an answer which someone might actually use in a production system, we can try a replacement approach:

public int countCode(String str) {
    return (str.length() - str.replace("code", "").length()) / 4;
}

Answer 3

Assume String is length 10 . When I goes from 0 to < 10 ie 0 to 9 , the test str.charAt(i+3)=='e' will cause i + 3 to exceed the length of the string when i >= 7 , and throw an exception. By limiting i to 3 less than the length , the loop will terminate before the index goes out of bounds.

Regarding your solution, I would offer the following alternative.

• split("co.e",-1) will split on the word co.e where . matches any character. The -1 will ensure trailing empty strings will be preserved (in case the string ends with codecodecode So the number array size would be 1 + the number of delimiters encountered so subtracting one is required.

public static int countCode(String str) {
        return (int)str.split("co.e",-1).length-1;
}

Since split takes a regex, either code or co.e can be used.

Updated

Better still would be to use Andy Turner's suggestion and increment do i += 3 when do code++ count.