compare to int java

compare string integer java

/**
    * Similar to compareTo method But compareTo doesn't return correct result for string+integer strings something like `A11` and `A9`
*/

private int newCompareTo(String comp1, String comp2) {
  // If any value has 0 length it means other value is bigger
  if (comp1.length() == 0) {
    if (comp2.length() == 0) {
      return 0;
    }
    return -1;
  } else if (comp2.length() == 0) {
    return 1;
  }
  if (TextUtils.isDigitsOnly(comp1)) {
    int val1 = Integer.parseInt(comp1);
    if (TextUtils.isDigitsOnly(comp2)) {
      int val2 = Integer.parseInt(comp2);
      return Integer.compare(val1, val2);
    } else {
      return comp1.compareTo(comp2);
    }

  } else {

    int minVal = Math.min(comp1.length(), comp2.length()), sameCount = 0;

    // Loop through two strings and check how many strings are same
    for (int i = 0;i < minVal;i++) {
      char leftVal = comp1.charAt(i), rightVal = comp2.charAt(i);
      if (leftVal == rightVal) {
        sameCount++;
      } else {
        break;
      }
    }
    if (sameCount == 0) {
      return comp1.compareTo(comp2);
    } else {
      String newStr1 = comp1.substring(sameCount), newStr2 = comp2.substring(sameCount);
      if (TextUtils.isDigitsOnly(newStr1) && TextUtils.isDigitsOnly(newStr2)) {
        return Integer.compare(Integer.parseInt(newStr1), Integer.parseInt(newStr2));
      } else {
        return comp1.compareTo(comp2);
      }
    }
  }
}

Posted by: Guest on November-21-2019

num == Integer.parseInt(str) is going to faster than str.equals("" + num) str.equals("" + num) will first convert num to string which is O(n) where n being the number of digits in the number. Then it will do a string concatenation again O(n) and then finally do the string comparison. String comparison in this case will be another O(n) - n being the number of digits in the number. So in all ~3*O(n) num == Integer.parseInt(str) will convert the string to integer which is O(n) again where n being the number of digits in the number. And then integer comparison is O(1). So just ~1*O(n) To summarize both are O(n) - but str.equals("" + num) has a higher constant and so is slower.

Collections.sort(data, new Comparator<TXCartData>() { @Override public int compare(TXCartData lhs, TXCartData rhs) { int n1=Integer.parseInt(lhs.rowId); int n2=Integer.parseInt(rhs.rowId); if (n1>=n2){ return 1; } return -1; } });

Code answers related to "compare to int java"

Code answers related to "Java"

Browse Popular Code Answers by Language

Answers for "compare to int java"

Code answers related to "compare to int java"

Code answers related to "Java"

Java Answers by Framework

Browse Popular Code Answers by Language

Popular Programming Languages

Advertisements

Company

Compilers

Help

Connect with us