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:

1. Whitespace: Ignore any leading whitespace (" ").
2. Signedness: Determine the sign by checking if the next character is '-' or '+', assuming positivity if neither present.
3. 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.
4. 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

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 $O(1)$.

Python3JavaGoJavaScriptC#PHP

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
            # 仅包含空格
            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:
            # 非数字，跳出循环体
            if not s[i].isdigit():
                break
            c = int(s[i])
            # 溢出判断
            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) == ' ') {
            // 仅包含空格
            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) {
            // 非数字，跳出循环体
            if (s.charAt(i) < '0' || s.charAt(i) > '9') break;
            // 溢出判断
            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;
    }
}

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;
};

// https://leetcode.com/problems/string-to-integer-atoi/

public partial 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);
    }
}

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