Class Solution
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'.'
.
To solve the String to Integer (atoi) problem in Java using a Solution
class, we’ll follow these steps:
- Define a
Solution
class with a method namedmyAtoi
. - Trim leading whitespace from the input string
s
. - Check if the string is empty after trimming. If so, return 0.
- Initialize variables to keep track of the sign of the integer (
sign
), the starting index of the numeric characters (start
), and the result (result
). - Check if the first character of the trimmed string is
'-'
or'+'
. Updatesign
accordingly, and move the starting index accordingly. - Iterate through the characters of the trimmed string starting from the
start
index:- Check if the current character is a digit. If not, break the loop.
- Convert the character to its numeric value and update the
result
. - Check if the result exceeds the 32-bit integer range. If so, clamp it to the range and return.
- Multiply the
result
by the sign and return the final value. - Handle edge cases where the input string is empty, consists of only whitespace, or contains non-numeric characters at the beginning.
Here’s the implementation:
public class Solution {
public int myAtoi(String s) {
s = s.trim();
if (s.isEmpty())
return 0;
int sign = 1;
int start = 0;
long result = 0;
if (s.charAt(0) == '-' || s.charAt(0) == '+') {
sign = (s.charAt(0) == '-') ? -1 : 1;
start++;
}
for (int i = start; i < s.length(); i++) {
char c = s.charAt(i);
if (!Character.isDigit(c))
break;
result = result * 10 + (c - '0');
if (result * sign > Integer.MAX_VALUE)
return Integer.MAX_VALUE;
if (result * sign < Integer.MIN_VALUE)
return Integer.MIN_VALUE;
}
return (int) (result * sign);
}
public static void main(String[] args) {
Solution solution = new Solution();
// Test cases
String s1 = "42";
System.out.println("Example 1 Output: " + solution.myAtoi(s1));
String s2 = " -42";
System.out.println("Example 2 Output: " + solution.myAtoi(s2));
String s3 = "4193 with words";
System.out.println("Example 3 Output: " + solution.myAtoi(s3));
String s4 = "words and 987";
System.out.println("Example 4 Output: " + solution.myAtoi(s4));
String s5 = "-91283472332";
System.out.println("Example 5 Output: " + solution.myAtoi(s5));
}
}
This implementation provides a solution to the String to Integer (atoi) problem in Java.
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Constructor Summary
Constructors -
Method Summary
Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
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Constructor Details
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Solution
public Solution()
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Method Details
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myAtoi
public int myAtoi(java.lang.String str)
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