Medium
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 is0
. Change the sign as necessary (from step 2). - 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 than231 - 1
should be clamped to231 - 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 [-231, 231 - 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 [-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.
Example 4:
Input: s = "words and 987"
Output: 0
Explanation:
Step 1: "words and 987" (no characters read because there is no leading whitespace)
^
Step 2: "words and 987" (no characters read because there is neither a '-' nor '+')
^
Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w')
^
The parsed integer is 0 because no digits were read.
Since 0 is in the range [-231, 231 - 1], the final result is 0.
Example 5:
Input: s = "-91283472332"
Output: -2147483648
Explanation:
Step 1: "-91283472332" (no characters read because there is no leading whitespace)
^
Step 2: "-91283472332" ('-' is read, so the result should be negative)
^
Step 3: "-91283472332" ("91283472332" is read in)
^
The parsed integer is -91283472332.
Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.
Constraints:
0 <= s.length <= 200
s
consists of English letters (lower-case and upper-case), digits (0-9
),' '
,'+'
,'-'
, and'.'
.
#include <stdio.h>
#include <limits.h>
#include <ctype.h>
int myAtoi(const char* str) {
if (str == NULL || *str == '\0') {
return 0;
}
int i = 0;
int num = 0;
int negativeSign = 0;
// Skip leading whitespace
while (str[i] == ' ') {
i++;
}
// Check for sign
if (str[i] == '+') {
i++;
} else if (str[i] == '-') {
i++;
negativeSign = 1;
}
// Convert string to integer
while (str[i] >= '0' && str[i] <= '9') {
int digit = str[i] - '0';
if (negativeSign) {
digit = -digit;
}
// Handle overflow and underflow
if (num < INT_MIN / 10 || (num == INT_MIN / 10 && digit < -8)) {
return INT_MIN;
}
if (num > INT_MAX / 10 || (num == INT_MAX / 10 && digit > 7)) {
return INT_MAX;
}
num = num * 10 + digit;
i++;
}
return num;
}