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MEDIUM

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:

1. Read in and ignore any leading whitespace.
2. Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is present. This determines if the final result is negative or positive. Assume the result is positive if neither is present.
3. 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.
4. 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).
5. If the integer is out of the 32-bit signed integer range [-2³¹, 2³¹ - 1], 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.
6. Return the integer as the final result.

Example

Input:

s = "42"

Output:

42

Input:

s = "   -42"

Output:

-42

Input:

s = "4193 with words"

Output:

4193

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Constraints

• 0 ≤ s.length ≤ 200
• s consists of English letters (lower-case and upper-case), digits, ' ', '+', '-', and '.'.

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Solution: Simulation

• Time Complexity: O(n), where n is the length of the string.
• Space Complexity: O(1), as only a constant amount of extra space is used.

C++

class Solution {
public:
    int myAtoi(string s) {
        int n = s.size(), ans = 0, sign = 1, i = 0;
        while(i < n && s[i] == ' ') i++;
        if(i < n && (s[i] == '+' || s[i] == '-')) sign = (s[i++] == '+') ? 1 : -1;
        while(i < n && isdigit(s[i])) {
            int cur = s[i++] - '0';
            // cur + ans*10 > INT_MAX
            if (ans > (INT_MAX-cur)/10) return sign == 1 ? INT_MAX : INT_MIN;
            ans = ans*10 + cur;
        }
        return ans*sign;
    }
};