LeetCode – String to Integer (atoi) (Java)

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Analysis

The following cases should be considered for this problem:

1. null or empty string
2. white spaces
3. +/- sign
4. calculate real value
5. handle min & max

Java Solution

public int atoi(String str) {
	if (str == null || str.length() < 1)
		return 0;
 
	// trim white spaces
	str = str.trim();
 
	char flag = '+';
 
	// check negative or positive
	int i = 0;
	if (str.charAt(0) == '-') {
		flag = '-';
		i++;
	} else if (str.charAt(0) == '+') {
		i++;
	}
	// use double to store result
	double result = 0;
 
	// calculate value
	while (str.length() > i && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
		result = result * 10 + (str.charAt(i) - '0');
		i++;
	}
 
	if (flag == '-')
		result = -result;
 
	// handle max and min
	if (result > Integer.MAX_VALUE)
		return Integer.MAX_VALUE;
 
	if (result < Integer.MIN_VALUE)
		return Integer.MIN_VALUE;
 
	return (int) result;
}

Category >> Algorithms

If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:

<pre><code> 
String foo = "bar";
</code></pre>

Rahul Jaiswal

Above code will fail for ” “.

Correct solution:

public int myAtoi(String str) { if(str == null) return 0; str = str.trim(); if(str.length() < 1) return 0;
boolean isNeg = false; int i = 0; if(str.charAt(0) == '-') { isNeg = true; i++; } else if(str.charAt(0) == '+') i++; double result = 0; while(i = '0' && str.charAt(i) Integer.MAX_VALUE) return Integer.MAX_VALUE;
if (result < Integer.MIN_VALUE) return Integer.MIN_VALUE; return (int)result; }
Bonsai

Integer has range [-2147483648 , 2147483647] in Java. If the number is -214748364, the algorithm processes it as a positive integer and then change sign. The number would exceed the integer range if an integer is used.
KARAN VERMA

One minor modification for prevent stackoverflowException: length check for string if(str.length()>0) {

if (str == null || str.length() 0) {
if (str.charAt(0) == '-') { flag = '-'; i++; } else if (str.charAt(0) == '+') { i++; } } // use double to store result double result = 0; // calculate value while (str.length() > i && str.charAt(i) >= '0' && str.charAt(i) Integer.MAX_VALUE) return Integer.MAX_VALUE; if (result < Integer.MIN_VALUE) return Integer.MIN_VALUE;
return (int) result;
Npy

in next iteration when we add new integer it should come at unit’s place.
kavya tg

Hi article is nice, can we parsing String to int in java 8 ?
i was searching for the information then i found very good information here, it best way to convert string to int i hope this may be use full.
Ahmed Saad

this is my solution to convert any string to double including fractions and + – signs

private static double checkString(String text) {
boolean isFraction = false; int fractionIndex = -1; boolean isSigned = false; boolean isPostive = true; if (text.charAt(0) == '+') { isSigned = true; isPostive = true; } else if (text.charAt(0) == '-') { isSigned = true; isPostive = false; } for (int i = 0; i 57 || currentChar < 48)) { return -1; } } int start = (isSigned) ? 1 : 0; double num; if (!isFraction) { return getNumber(text, start, text.length() - 1, false); } else { if (fractionIndex - start = start; i--) { char currentChar = text.charAt(i); num = num + level * (currentChar - 48); level = level * 10; } if (isFraction) { num = num / (level); }
return num; }
Brenda Martis

Why was double used for result?
Brandon Donnelson

Alternative solution.

public int myAtoi(String str) { if (str == null || str.isEmpty()) { return 0; }
str = str.trim(); if (str.equals("-")) { return 0; } if (str.equals("+")) { return 0; } int signAdd = 0; boolean sign = false; long result = 0; for (int i = 0; i 1) { return 0; } } if (c == '-') { sign = true; } else if (c == '+') { sign = false; } else if (c >= '0' && c Integer.MAX_VALUE) { return Integer.MAX_VALUE; } else if (sign && result *-1 < Integer.MIN_VALUE) { return Integer.MIN_VALUE; } } if (sign) { result *= -1; }
return (int) result; }
Sékine Coulibaly

I guess it mean for positive numbers, eg. 42.
Obviously 42 is positive, and equals to +42, but there is no leading “+” sign.
loyolakarl

kj;ljdaLJD
I am writing some code as said before lol. :)
Rory

Do not need this part

// handle max and min

if (result > Integer.MAX_VALUE)

return Integer.MAX_VALUE;

if (result < Integer.MIN_VALUE)

return Integer.MIN_VALUE;
Renzo

Solution in 11 lines of Python:https://github.com/renzon/code_interview_training/blob/master/string_to_int.py
The good thing is that for integer I don’t have to worry once they are arbitrary large in Python (not arbitrary, but as long as it fit in your memory)
joseph

Really should check the string length after doing the trim(), in case it was a string of just white spaces.
Peng

I think the overflow check should be placed within for loop to attach this problem
Arnaud Hebert

Why are we multiplying by 10: result = result * 10 ?
lily

well,if the str is very very very big,such as it is bigger than long,how can you handle this?
Taoufik Bdiri

he could use :
Character.getNumericValue(str.charAt(i))
Chris

Why the sign is not always supposed to be at the beginning of the string? Please give a example? Thank you.
sebastian

hi is doing that because charAt return a char so you can’t sum a char, you need a way to convert to integer one choice is that and only is charAt(pos) – 48

Bye.
armandm

why are you substracting ” to the char
?
uee

The sign is not always supposed to be at the beginning of the string. And you also didn’t consider the situation that other character show in the string.

LeetCode – String to Integer (atoi) (Java)

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