String to Integer (atoi)

Problem#

Implement a parser that converts a string to a 32-bit signed integer (clamp to [-2^31, 2^31-1] on overflow). Skip leading whitespace, parse optional sign, consume the digit run.

Examples#

"42" → 42.
" -42" → -42; "4193 with words" → 4193; "words and 987" → 0.

Constraints#

0 <= s.length <= 200.

Approach#

Skip whitespace. Consume optional sign. Parse digits, clamping at each step before multiplying by 10.

Solution#

class Solution {
public:
    int myAtoi(string s) {
        int i = 0, n = s.size();
        while (i < n && s[i] == ' ') ++i;
        int sign = 1;
        if (i < n && (s[i] == '+' || s[i] == '-')) {
            if (s[i] == '-') sign = -1;
            ++i;
        }
        long long ans = 0;
        while (i < n && isdigit(s[i])) {
            ans = ans * 10 + (s[i] - '0');
            if (sign * ans > INT_MAX) return INT_MAX;
            if (sign * ans < INT_MIN) return INT_MIN;
            ++i;
        }
        return (int)(sign * ans);
    }
};

import "math"

func myAtoi(s string) int {
    i, n := 0, len(s)
    for i < n && s[i] == ' ' {
        i++
    }
    sign := 1
    if i < n && (s[i] == '+' || s[i] == '-') {
        if s[i] == '-' {
            sign = -1
        }
        i++
    }
    var ans int64
    for i < n && s[i] >= '0' && s[i] <= '9' {
        ans = ans*10 + int64(s[i]-'0')
        if int64(sign)*ans > math.MaxInt32 {
            return math.MaxInt32
        }
        if int64(sign)*ans < math.MinInt32 {
            return math.MinInt32
        }
        i++
    }
    return int(int64(sign) * ans)
}

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

function myAtoi(s) {
    const INT_MAX = 2147483647;
    const INT_MIN = -2147483648;
    let i = 0;
    const n = s.length;
    while (i < n && s[i] === ' ') i++;
    let sign = 1;
    if (i < n && (s[i] === '+' || s[i] === '-')) {
        if (s[i] === '-') sign = -1;
        i++;
    }
    let ans = 0;
    while (i < n && s[i] >= '0' && s[i] <= '9') {
        ans = ans * 10 + (s.charCodeAt(i) - 48);
        if (sign * ans > INT_MAX) return INT_MAX;
        if (sign * ans < INT_MIN) return INT_MIN;
        i++;
    }
    return sign * ans;
}

class Solution {
    public int myAtoi(String s) {
        int i = 0, n = s.length();
        while (i < n && s.charAt(i) == ' ') i++;
        int sign = 1;
        if (i < n && (s.charAt(i) == '+' || s.charAt(i) == '-')) {
            if (s.charAt(i) == '-') sign = -1;
            i++;
        }
        long ans = 0;
        while (i < n && Character.isDigit(s.charAt(i))) {
            ans = ans * 10 + (s.charAt(i) - '0');
            if (sign * ans > Integer.MAX_VALUE) return Integer.MAX_VALUE;
            if (sign * ans < Integer.MIN_VALUE) return Integer.MIN_VALUE;
            i++;
        }
        return (int) (sign * ans);
    }
}

function myAtoi(s: string): number {
    const INT_MAX = 2147483647;
    const INT_MIN = -2147483648;
    let i = 0;
    const n = s.length;
    while (i < n && s[i] === ' ') i++;
    let sign = 1;
    if (i < n && (s[i] === '+' || s[i] === '-')) {
        if (s[i] === '-') sign = -1;
        i++;
    }
    let ans = 0;
    while (i < n && s[i] >= '0' && s[i] <= '9') {
        ans = ans * 10 + (s.charCodeAt(i) - 48);
        if (sign * ans > INT_MAX) return INT_MAX;
        if (sign * ans < INT_MIN) return INT_MIN;
        i++;
    }
    return sign * ans;
}

Editorial#

Using long long for the accumulator lets us multiply by 10 and add a digit without immediate overflow; the per-step clamp catches any value beyond int range as it’s being built. Whitespace and sign are each parsed at most once.

Complexity#

Time: O(n).
Space: O(1).

Problem#

Examples#

Constraints#

Approach#

Solution#

Editorial#

Complexity#

Concept revision#

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