Handle overflow: The result must be within the range of a 32-bit signed integer [-2^31, 2^31 - 1], which is [-2147483648, 2147483647]. If the number would overflow: ... The solution needs to carefully handle edge cases like empty strings, strings with only spaces, and integer overflow situations ...

8. String to Integer (atoi) Medium · 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: Whitespace: Ignore any leading whitespace (" "). Signedness: ...

October 27, 2023 - 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).

LeetCode Solutions in C++23, Java, Python, MySQL, and TypeScript.

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. Rounding: If the integer is out of the 32-bit signed integer ...

September 12, 2024 - class Solution { public int myAtoi(String s) { s = s.strip(); if (s.isEmpty()) return 0; final int sign = s.charAt(0) == '-' ? -1 : 1; if (s.charAt(0) == '+' || s.charAt(0) == '-') s = s.substring(1); long num = 0; for (final char c : s.toCharArray()) { if (!Character.isDigit(c)) break; num = num * 10 + (c - '0'); if (sign * num <= Integer.MIN_VALUE) return Integer.MIN_VALUE; if (sign * num >= Integer.MAX_VALUE) return Integer.MAX_VALUE; } return sign * (int) num; } } // code provided by PROGIEZ

February 28, 2025 - Problem8. String to Integer (atoi) Java1234567891011121314151617181920212223242526272829303132333435363738394041class Solution { public int myAtoi(String s) { int re = 0; lo

public int myAtoi(String str) { if (str.isEmpty()) return 0; int sign = 1, base = 0, i = 0, n = str.length(); while (i < n && str.charAt(i) == ' ') ++i; if (i < n && (str.charAt(i) == '+' || str.charAt(i) == '-')) { sign = (str.charAt(i++) == '+') ? 1 : -1; } while (i < n && str.charAt(i) >= '0' && str.charAt(i) <= '9') { if (base > Integer.MAX_VALUE / 10 || (base == Integer.MAX_VALUE / 10 && str.charAt(i) - '0' > 7)) { return (sign == 1) ? Integer.MAX_VALUE : Integer.MIN_VALUE; } base = 10 * base + (str.charAt(i++) - '0'); } return base * sign; } http://www.jiuzhang.com/solutions/string-to-in

July 17, 2024 - Read each character until a non-digit character is encountered or the end of the string is reached. Convert these characters into a number. ... If the number is outside the range of a 32-bit signed integer, clamp it to the range limits.

Integer.MAX_VALUE : Integer.MIN_VALUE; is necessary and enough to test all the overflow, 2137385647, -2147483647, 2137385648(return 2137385647) and -2147483648 will all return the right answers. public int atoi(String str) { if(str==null || str.length()==0) return 0; int start=0; int res=0; boolean pos=true; while(start<str.length() && str.charAt(start)==' ') start++; if(start==str.length()) return 0; if(str.charAt(start)=='+'||str.charAt(start)=='-') { pos=str.charAt(start++)=='+'; } for(int i=start;i<str.length() && str.charAt(i)<='9'&&str.charAt(i)>='0';i++){ int dig=str.charAt(i)-'0'; if((Integer.MAX_VALUE-dig)/10<res) return (pos)?

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function. If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

* @param str: A string · * @return An integer · */ · public int atoi(String str) { · if (str == null || str.length() == 0) { · return 0; · } · str = str.replaceAll("\\s",""); · if (!str.matches("[+-]?(?:\\d+(?:\\.\\d*)?|\\.\\d+)")) { ·

if (ans > Integer.MAX_VALUE / 10 || (ans == Integer.MAX_VALUE / 10 && str.charAt(i) > '7')) {

public int myAtoi(String s) { long rst = 0; long rst_too_large = (long)Integer.MAX_VALUE + 1; boolean negative = false; boolean found_digits = false; for (int i=0; i<s.length(); ++i) { // ch: +/-, digit, '.', space, oth ·

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function. If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.