# 8. String to Integer (atoi)

## Description

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi function).

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

2. Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
3. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
4. Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
5. If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.
6. Return the integer as the final result.

Note:

• Only the space character ' ' is considered a whitespace character.
• Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.

Example 1:

Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, 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)
^
The parsed integer is 42.
Since 42 is in the range [-231, 231 - 1], the final result is 42.


Example 2:

Input: s = "   -42"
Output: -42
Explanation:
^
Step 2: "   -42" ('-' is read, so the result should be negative)
^
Step 3: "   -42" ("42" is read in)
^
The parsed integer is -42.
Since -42 is in the range [-231, 231 - 1], the final result is -42.


Example 3:

Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
^
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
^
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
^
The parsed integer is 4193.
Since 4193 is in the range [-231, 231 - 1], the final result is 4193.


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 $O(1)$.

• 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 (int j = i; j < n; ++j) {
if (s.charAt(j) < '0' || s.charAt(j) > '9') break;
if (res > flag || (res == flag && s.charAt(j) > '7'))
return sign > 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
res = res * 10 + (s.charAt(j) - '0');
}
return sign * res;
}
}

• 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
# only contains blank space
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:
# not a number, exit the loop
if not s[i].isdigit():
break
c = int(s[i])
# if overflows
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


• 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);
}
}