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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:
'-' 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."123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).[-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.Note:
' ' is considered a whitespace character.
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.
Constraints:
0 <= s.length <= 200s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.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:
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
# 仅包含空格
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:
# 非数字，跳出循环体
if not s[i].isdigit():
break
c = int(s[i])
# 溢出判断
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
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) == ' ') {
// 仅包含空格
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 (; i < n; ++i) {
// 非数字，跳出循环体
if (s.charAt(i) < '0' || s.charAt(i) > '9') break;
// 溢出判断
if (res > flag || (res == flag && s.charAt(i) > '7'))
return sign > 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
res = res * 10 + (s.charAt(i) - '0');
}
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);
}
}
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