String to Integer

8 String to Integer

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

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.

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

String to Integer

Note:

Only the space character ‘ ‘ is considered as whitespace character.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

Example 1:

Input: "42"
Output: 42
Example 2:

Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.
Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.

Problem Analysis

Parse the conditions:

1. Discard whitespace until we faced the first non-whitespaces characters. (A string might contians all white spcaes)
2. Prase +/- for the first characters and assign to sign variable.
3. Start to look at the number.
4. -1. Meet invalid character
5. -2. Meet 0, and we should skip all 0 as the beginning of number
6. -3. Meet number and we start to calculate the number
7. Parse the number:
8. -1. Check if we meet invlaid character inside the number
9. -2. Check if the current number might exceed largest or smallest integer. (Integer.MAX_VALUE - digit) / 10 < sum)
10. -3. Keep adding the numebr into sum variable
11. Return the value with correct sign.

Algorithm Analysis

Just following the instructions to construct this algorithm.

Time Complexity Analysis

• Time: O(n)

Code Implementation

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79

class Solution 
{
    public int myAtoi(String str) 
    {
        int sign = 1;
        int sum = 0;
        int index = 0;
        
        /* Empty String */
        if(str.length() == 0) 
        {
            return sum;
        }
            
        /* Discard White Space */
        while(index < str.length() && str.charAt(index) == ' ') 
        {
            index ++;
        }
        
        /* All string are white spaces */
        if(index == str.length()) 
        {
            return sum;
        }
        
        /* Currently, there is no space before the first valid character */
        /* Check if I need to give a value to sign or meet invalid number */
        if(str.charAt(index) == '+' ) 
        {
            sign = 1;
            index++;
        } 
        else if(str.charAt(index) == '-') 
        {
            sign = -1;
            index ++;
        } 
        else if(str.charAt(index) == '0')
        {
            /* Remove all 0 */
            while(index < str.length() && str.charAt(index) == '0') 
            {
                index++;
            }
            /* All valid string contains 0 */
            if(index == str.length()) {
                return sum;
            }
        }
        else if(str.charAt(index) > '9' || str.charAt(index) < '1') 
        {
            /* Meet invalid character */
            return sum;
        }

        /* Prase the number */
        for(int i = index; i < str.length(); i++) 
        {
            /* Meet invalid character */
            if(str.charAt(i) > '9' || str.charAt(i) < '0') 
            {
                return sign * sum;
            }
            
            int digit = str.charAt(i) - '0';
            
            /* Check if the current sum will exceed limitation */
            if((Integer.MAX_VALUE - digit) / 10 < sum) 
            {
                return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }

            sum *= 10;
            sum += digit;
        }
        return sign * sum;    
    }
}

• ← Previous Post
• Next Post →