Problem:
Given 2 non-negative ints, a and b, return their sum, so long as the sum has the same number of digits as a. If the sum has more digits than a, just return a without b. (Note: one way to compute the number of digits of a non-negative int n is to convert it to a string with String.valueOf(n) and then check the length of the string.)
sumLimit(2, 3) → 5
sumLimit(8, 3) → 8
sumLimit(8, 1) → 9
Solution:
public int sumLimit(int a, int b) {
String str = Integer.toString(a+b);
String str2 = Integer.toString(a);
if(str.length() == str2.length())
return a+b;
else
return a;
}
boolean isEqual = String.valueOf(a).length() == String.valueOf(a+b).length();
ReplyDeletereturn isEqual ? a + b : a;
if (String.valueOf(a+b).length() == String.valueOf(a).length())
ReplyDeletereturn a + b;
return a;
public static int sumLimit2(int a, int b) {
ReplyDeleteString str = Integer.toString(a + b);
String str2 = Integer.toString(a);
if (str.length() == str2.length()) {
return a + b;
}
return a;
}
public int sumLimit(int a, int b) {
ReplyDeleteint sum = a + b;
if ( no_of_digits(sum) > no_of_digits(a) ) return a;
return sum;
}
public int no_of_digits(int number) {
return String.valueOf(number).length();
}
return (String.valueOf(a+b).length() == String.valueOf(a).length()) ? a+b:a;
ReplyDeleteIt's simple, but one liners (when possible) will always be more pleasant to read through.