String to Integer (atoi)
String to Integer (atoi):
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:
Read in and ignore any leading whitespace.
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.
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.
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).
If the integer is out of the 32-bit signed integer range [-2^31, 2^31 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -2^31 should be clamped to -2^31, and integers greater than 2^31 - 1 should be clamped to 2^31 - 1.
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 [-2^31, 2^31 - 1], the final result is 42.
Example 2:
Input: s = " -42"
Output: -42
Explanation:
Step 1: " -42" (leading whitespace is read and ignored)
^
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 [-2^31, 2^31 - 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 [-2^31, 2^31 - 1], the final result is 4193.
Constraints:
0 <= s.length <= 200sconsists of English letters (lower-case and upper-case), digits (0-9),' ','+','-', and'.'.
Try this Problem on your own or check similar problems:
Solution:
- Java
- JavaScript
- Python
- C++
public int myAtoi(String s) {
int i = 0;
boolean isNegative = false, clamp = false;
int number = 0;
Character prev = null;
while(i < s.length() && !Character.isDigit(s.charAt(i))){
if(s.charAt(i) == '-' && (i + 1 < s.length() && Character.isDigit(s.charAt(i+1)))) isNegative = true;
if((Character.isLetter(s.charAt(i)) || s.charAt(i) == '.') ||
(s.charAt(i) == '-' && prev != null && prev == '+') ||
(s.charAt(i) == '+' && prev != null && prev == '-') ||
(prev != null && (prev == '+' || prev == '-'))) return 0;
prev = s.charAt(i);
++i;
}
while(i < s.length() && Character.isDigit(s.charAt(i))){
if((long) number * 10 > Integer.MAX_VALUE) clamp = true;
number *= 10;
int digit = s.charAt(i) - '0';
if((long)number + digit > Integer.MAX_VALUE) clamp = true;
number += s.charAt(i) - '0';
++i;
}
if(clamp && isNegative) return Integer.MIN_VALUE;
else if(clamp) return Integer.MAX_VALUE;
if(isNegative){
number *= -1;
}
return number;
}
/**
* @param {string} s
* @return {number}
*/
var myAtoi = function (s) {
const MIN_VALUE = Math.pow(-2, 31);
const MAX_VALUE = Math.pow(2, 31) - 1;
const result = Number(s.trimLeft().match(/^[-\+]?\d+/));
if (result < MIN_VALUE) return MIN_VALUE;
if (result > MAX_VALUE) return MAX_VALUE;
return result;
};
class Solution:
def myAtoi(self, string: str) -> int:
stripped = string.strip()
sign = 1
if not stripped:
return 0
elif stripped[0] in ("+", "-"):
if stripped[0] == "-":
sign = -1
stripped = stripped[1:]
elif not stripped[0].isdigit():
return 0
if not stripped:
return 0
try:
ans = int(stripped[0])
stripped = stripped[1:]
for c in stripped:
if c.isdigit():
ans = ans * 10 + int(c)
else:
break
if sign == 1:
return sign * ans if ans < 2**31 else 2147483647
return -ans if ans <= 2**31 else -2147483648
except:
return 0
class Solution {
public:
int myAtoi(string s) {
int len = s.size();
double num = 0;
int i=0;
while(s[i] == ' '){
i++;
}
bool positive = s[i] == '+';
bool negative = s[i] == '-';
positive == true ? i++ : i;
negative == true ? i++ : i;
while(i < len && s[i] >= '0' && s[i] <= '9'){
num = num*10 + (s[i]-'0');
i++;
}
num = negative ? -num : num;
num = (num > INT_MAX) ? INT_MAX : num;
num = (num < INT_MIN) ? INT_MIN : num;
return int(num);
}
};
Time/Space Complexity:
- Time Complexity: O(n)
- Space Complexity: O(1)
Explanation:
We have linear time complexity since we traverse the whole string. There is nothing special to the algorithm besides handling the invalid cases and checking if we have overflow with the current addition/multiplication to set the clamp (determines if we have to clamp the number to int max and min value). We first traverse the string until we hit the first number, but before that we must check for invalid arrangement (where we return 0). We have the following cases:
- We first hit a non-digit and no space character
- There is a
+/-character followed by the opposite sign-/+ - We have
+/-sign followed by non-digit character
We also check if we have valid negative number by setting the flag isNegative. If we still have a valid string, we create the number from it while also checking for overflow. We either return the clamped version, negative version (if number is negative) or just raw number value.