MediumStringsMath

String to Integer (atoi)

1 approach, code in every language

Convert a string into a signed 32-bit integer by scanning it the way a hand-written parser would. First skip any leading spaces, then read an optional single plus or minus sign, then collect the run of decimal digits that follows. Reading stops at the first character that is not a digit, and everything after that point is ignored.

If no digits appear the result is 0. When the assembled number falls outside the signed 32-bit range, clamp it: values below the minimum become -2^31 and values above the maximum become 2^31 - 1.

Example 1

Input: s = " -042"

Output: -42

Leading spaces are skipped, the minus sign is recorded, and the leading zero in 042 is absorbed to give -42.

Example 2

Input: s = "4193 with words"

Output: 4193

Digits are gathered until the space, so the trailing text is discarded.

Constraints

0 <= s.length <= 200
s consists of spaces, digits, letters, '+', '-', and '.'
The final value is clamped to the range [-2^31, 2^31 - 1]

Single Linear Scan with Clamping

Walk the string once through four phases (whitespace, sign, digits, stop) and clamp the accumulated value to the 32-bit range.

Loading animation…

Intuition

The rules describe a small state machine, so a single left-to-right pass is enough. First we drop leading spaces, then peek at one optional sign character, then keep folding digits into an accumulator with value = value * 10 + digit. As soon as a non-digit shows up we stop, because the specification says the parse ends there. Overflow is handled by comparing against the 32-bit boundaries as we build the number and clamping to the nearest limit rather than letting the value wrap around.

Algorithm walkthrough

1Advance an index past every leading space character.
2If the current character is '+' or '-', note the sign and move one step forward.
3While the current character is a digit, multiply the running value by ten and add the digit.
4During accumulation, if the value crosses the 32-bit limit, immediately return the clamped minimum or maximum.
5When a non-digit is reached or the string ends, apply the sign and return the value.

Implementation

class Solution:
    def myAtoi(self, s: str) -> int:
        INT_MIN, INT_MAX = -2**31, 2**31 - 1
        i, n = 0, len(s)
        while i < n and s[i] == ' ':
            i += 1
        sign = 1
        if i < n and (s[i] == '+' or s[i] == '-'):
            if s[i] == '-':
                sign = -1
            i += 1
        result = 0
        while i < n and s[i].isdigit():
            result = result * 10 + int(s[i])
            i += 1
        result *= sign
        if result < INT_MIN:
            return INT_MIN
        if result > INT_MAX:
            return INT_MAX
        return result

Time

O(n)

Space

O(1)

One pass over the string with a fixed set of scalar variables.

Create a free account to see the solution

You get the step-by-step animation, the intuition and clean code in every language. Free, no credit card.

Create a free account Already have an account? Sign in

Back to Strings

After the pattern

Now solve one you haven't seen.

Salvage the Leading Number is our own problem on the technique behind String to Integer (atoi): a new statement, so there's no answer to remember. Write it in your browser and run it against graded tests.

Practise it with tests All practice problems

1,194 problems · 15,062 graded tests · free