8. String to Integer (atoi)

Description

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer.

The algorithm for myAtoi(string s) is as follows:

Whitespace: Ignore any leading whitespace (" ").
Signedness: Determine the sign by checking if the next character is '-' or '+', assuming positivity if neither present.
Conversion: Read the integer by skipping leading zeros until a non-digit character is encountered or the end of the string is reached. If no digits were read, then the result is 0.
Rounding: If the integer is out of the 32-bit signed integer range [-2³¹, 2³¹ - 1], then round the integer to remain in the range. Specifically, integers less than -2³¹ should be rounded to -2³¹, and integers greater than 2³¹ - 1 should be rounded to 2³¹ - 1.

Return the integer as the final result.

Example 1:

Input: s = "42"

Output: 42

Explanation:

The underlined characters are what is read in and the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
         ^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "42" ("42" is read in)
           ^

Example 2:

Input: s = " -042"

Output: -42

Explanation:

Step 1: "   -042" (leading whitespace is read and ignored)
            ^
Step 2: "   -042" ('-' is read, so the result should be negative)
             ^
Step 3: "   -042" ("042" is read in, leading zeros ignored in the result)
               ^

Example 3:

Input: s = "1337c0d3"

Output: 1337

Explanation:

Step 1: "1337c0d3" (no characters read because there is no leading whitespace)
         ^
Step 2: "1337c0d3" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "1337c0d3" ("1337" is read in; reading stops because the next character is a non-digit)
             ^

Example 4:

Input: s = "0-1"

Output: 0

Explanation:

Step 1: "0-1" (no characters read because there is no leading whitespace)
         ^
Step 2: "0-1" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "0-1" ("0" is read in; reading stops because the next character is a non-digit)
          ^

Example 5:

Input: s = "words and 987"

Output: 0

Explanation:

Reading stops at the first non-digit character 'w'.

Constraints:

0 <= s.length <= 200
s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.

Solutions

Solution 1: Traverse the String

Thinking

A single left-to-right scan is enough; \(n \le 200\) is not a complexity issue. The easy mistakes are the order of rules: skip leading spaces, consume at most one sign, then take consecutive digits, stop at the first non-digit, and clamp to 32 bits.

Without 64-bit integers we cannot finish the product and clip later. We accumulate the absolute value, comparing against \(\lfloor (2^{31}-1)/10 \rfloor\) before multiplying by \(10\), and saturate to \(2^{31}-1\) or \(-2^{31}\) on overflow.

Both signs share that upper bound; the sign is applied at the end. An empty string or all spaces has no digits, so we return \(0\).

First, we determine whether the string is empty. If it is, we directly return \(0\).

Otherwise, we need to traverse the string, skip the leading spaces, and determine whether the first non-space character is a positive or negative sign.

Then we traverse the following characters. If it is a digit, we judge whether adding this digit will cause integer overflow. If it does, we return the result according to the positive or negative sign. Otherwise, we add the digit to the result. We continue to traverse the following characters until we encounter a non-digit character or the traversal ends.

After the traversal ends, we return the result according to the positive or negative sign.

The time complexity is \(O(n)\), where \(n\) is the length of the string. We only need to process all characters in turn. The space complexity is \(O(1)\).

Python3JavaC++GoJavaScriptC#PHPC

class Solution:
    def myAtoi(self, s: str) -> int:
        if not s:
            return 0
        n = len(s)
        if n == 0:
            return 0
        i = 0
        while s[i] == ' ':
            i += 1
            if i == n:
                return 0
        sign = -1 if s[i] == '-' else 1
        if s[i] in ['-', '+']:
            i += 1
        res, flag = 0, (2**31 - 1) // 10
        while i < n:
            if not s[i].isdigit():
                break
            c = int(s[i])
            if res > flag or (res == flag and c > 7):
                return 2**31 - 1 if sign > 0 else -(2**31)
            res = res * 10 + c
            i += 1
        return sign * res

class Solution {
    public int myAtoi(String s) {
        if (s == null) return 0;
        int n = s.length();
        if (n == 0) return 0;
        int i = 0;
        while (s.charAt(i) == ' ') {
            if (++i == n) return 0;
        }
        int sign = 1;
        if (s.charAt(i) == '-') sign = -1;
        if (s.charAt(i) == '-' || s.charAt(i) == '+') ++i;
        int res = 0, flag = Integer.MAX_VALUE / 10;
        for (; i < n; ++i) {
            if (s.charAt(i) < '0' || s.charAt(i) > '9') break;
            if (res > flag || (res == flag && s.charAt(i) > '7'))
                return sign > 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            res = res * 10 + (s.charAt(i) - '0');
        }
        return sign * res;
    }
}

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] == '+')) {
            sign = s[i] == '-' ? -1 : 1;
            ++i;
        }

        int res = 0;
        while (i < n && isdigit(s[i])) {
            int digit = s[i] - '0';
            if (res > INT_MAX / 10 || (res == INT_MAX / 10 && digit > INT_MAX % 10)) {
                return sign == 1 ? INT_MAX : INT_MIN;
            }
            res = res * 10 + digit;
            ++i;
        }
        return res * sign;
    }
};

func myAtoi(s string) int {
    i, n := 0, len(s)
    num := 0

    for i < n && s[i] == ' ' {
        i++
    }
    if i == n {
        return 0
    }

    sign := 1
    if s[i] == '-' {
        sign = -1
        i++
    } else if s[i] == '+' {
        i++
    }

    for i < n && s[i] >= '0' && s[i] <= '9' {
        num = num*10 + int(s[i]-'0')
        i++
        if num > math.MaxInt32 {
            break
        }
    }

    if num > math.MaxInt32 {
        if sign == -1 {
            return math.MinInt32
        }
        return math.MaxInt32
    }
    return sign * num
}

const myAtoi = function (str) {
    str = str.trim();
    if (!str) return 0;
    let isPositive = 1;
    let i = 0,
        ans = 0;
    if (str[i] === '+') {
        isPositive = 1;
        i++;
    } else if (str[i] === '-') {
        isPositive = 0;
        i++;
    }
    for (; i < str.length; i++) {
        let t = str.charCodeAt(i) - 48;
        if (t > 9 || t < 0) break;
        if (ans > 2147483647 / 10 || ans > (2147483647 - t) / 10) {
            return isPositive ? 2147483647 : -2147483648;
        } else {
            ans = ans * 10 + t;
        }
    }
    return isPositive ? ans : -ans;
};

public class Solution {
    public int MyAtoi(string str) {
        int i = 0;
        long result = 0;
        bool minus = false;
        while (i < str.Length && char.IsWhiteSpace(str[i])) {
            ++i;
        }
        if (i < str.Length) {
            if (str[i] == '+') {
                ++i;
            }
            else if (str[i] == '-') {
                minus = true;
                ++i;
            }
        }
        while (i < str.Length && char.IsDigit(str[i])) {
            result = result * 10 + str[i] - '0';
            if (result > int.MaxValue) {
                break;
            }
            ++i;
        }
        if (minus) result = -result;
        if (result > int.MaxValue) {
            result = int.MaxValue;
        }
        if (result < int.MinValue) {
            result = int.MinValue;
        }
        return (int)result;
    }
}

class Solution {
    /**
     * @param string $s
     * @return int
     */

    function myAtoi($s) {
        $s = str_replace('e', 'x', $s);
        if (intval($s) < pow(-2, 31)) {
            return -2147483648;
        }
        if (intval($s) > pow(2, 31) - 1) {
            return 2147483647;
        }
        return intval($s);
    }
}

int myAtoi(char* s) {
    int i = 0;

    while (s[i] == ' ') {
        i++;
    }

    int sign = 1;
    if (s[i] == '-' || s[i] == '+') {
        sign = (s[i] == '-') ? -1 : 1;
        i++;
    }

    int res = 0;
    while (isdigit(s[i])) {
        int digit = s[i] - '0';
        if (res > INT_MAX / 10 || (res == INT_MAX / 10 && digit > INT_MAX % 10)) {
            return sign == 1 ? INT_MAX : INT_MIN;
        }
        res = res * 10 + digit;
        i++;
    }

    return res * sign;
}

8. String to Integer (atoi)

Description

Solutions

Solution 1: Traverse the String

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