Leetcode and Lintcode

String to Integer(atoi)

Implement function atoi to convert a string to an integer.

If no valid conversion could be performed, a zero value is returned.

If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

Example
"10" => 10

"-1" => -1

"123123123123123" => 2147483647

"1.0" => 1

Leetcode
Lintcode

Solution:

Detailed problem.
Three derails should be noted:

1. valid char c : c - '0',
2. not valid char: stop
3. '+' or '-' sign at the start, record it
4. Overflow: INT_MAX (2147483647) or INT_MIN (-2147483648) is returned

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public class Solution {
    public int atoi(String str) {
        // write your code here
        if (str == null || str.length() == 0)
            return 0;
        int sign = 1;
        int index = 0;

        // if we use long, we should return MAX or MIN value directly,
        double num = 0; 

        
        str = str.trim();

        if (str.charAt(index) == '+')
            index++;
        else if (str.charAt(index) == '-'){
            sign = -1;
            index++;
        }
        
        for (; index < str.length(); index++){
            char current = str.charAt(index);
            if (current > '9' || current < '0' || num > Integer.MAX_VALUE)
                break;
            num = num * 10 + (current - '0'); 
        }

        return (int) (sign * num);
    }


    // If double and long are not allowed
    public int atoi(String str) {
        // write your code here
        if (str == null || str.length() == 0)
            return 0;
        int sign = 1;
        int index = 0;
        int num = 0;
        
        str = str.trim();

        if (str.charAt(index) == '+')
            index++;
        else if (str.charAt(index) == '-'){
            sign = -1;
            index++;
        }
        
        for (; index < str.length(); index++){
            char current = str.charAt(index);
            if (current > '9' || current < '0') 
                break;
            if (num > Integer.MAX_VALUE / 10
              || (num == Integer.MAX_VALUE / 10 && num > Integer.MAX_VALUE % 10))
                return sign > 0? Integer.MAX_VALUE : Integer.MIN_VALUE;
            num = num * 10 + (current - '0'); 
        }

        return (int) (sign * num);
    }
}