LeetCode - String to Integer (atoi)

4 minute read

Problem description

description

Implement atoi which converts a string to an 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.

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.

Example 1:

1
2
Input: "42"
Output: 42

Example 2:

1
2
3
4
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:

1
2
3
Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

1
2
3
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.

Example 5:

1
2
3
4
Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−231) is returned.

Analysis

The main part of this problem is to deal with the edge cases. They are:

  • leading and trailing spaces

  • break when not a digit

  • is negative or not

  • overflow issue

some edge cases are not mentioned in the description (which I failed for these cases)

  • “+1” case indicate that should deal with first character is ‘+’

  • ” 0000000000012345678” case indicate that should deal with the leading 0

So Here’s the code:

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
public class StringtoInteger {
  public int myAtoi(String str) {
    str = str.trim();

    int res = 0;
    int factor = 0;
    int len = 0;
    for (char c : str.toCharArray()) {
      // deal with factor
      if (factor == 0) {
        if (c == '-') {
          factor = -1;
          continue;
        } else if (c == '+') {
          factor = 1;
          continue;
        } else if (Character.isDigit(c)) {
          factor = 1;
        } else {
          return 0;
        }
      }

      if (Character.isDigit(c)) {
        int v = c - '0';
        if (len > 9
            || (len == 9
                && (res * factor > Integer.MAX_VALUE / 10
                    || res * factor < Integer.MIN_VALUE / 10
                    || (res * factor == Integer.MAX_VALUE / 10 && v > 7)
                    || (res * factor == Integer.MIN_VALUE / 10 && v > 8)))) {
          return factor > 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
        }

        res = res * 10 + v;

        if (res != 0){
          len++;
        }
      } else {
        break;
      }
    }
    return factor * res;
  }
}

Something can be optimized in my code for the overflow part, which is line 26 to 33. we don’t need to consider the positive and negative case and differentiate MAX and MIN, because the only cases to break that rule is the MAX and MIN itself. So we can just check the smaller one which is the MAX.

Another thing is that we can use a index to check the factor and make the code more reasonable.

Here’s the sample code provided by LeetCode:

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
class Solution {
     public int myAtoi(String str) {
    //if empty return 0;
    int sign = 1;
    int digit;
    int total = 0;
    int idx = 0;
    //step1: clear all the empty spaces
    str = str.trim();
    if (str.length() == 0) {
      return 0;
    }
    //step2: check sign if the first char is sign
    if (str.charAt(idx) == '+') {
      sign = 1;
      idx++;
    } else if (str.charAt(idx) == '-') {
      sign = -1;
      idx++;
    }
    //step3: convert chars from idx 1 to last numeric
    char ch;
    for (; idx < str.length(); idx++) {
      ch = str.charAt(idx);
      if (ch < '0' || ch > '9') break;
      digit = ch - '0';
      if (total > Integer.MAX_VALUE / 10 || (total == Integer.MAX_VALUE / 10 && digit > Integer.MAX_VALUE % 10)) {
        return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
      }
      total = digit + (10 * total);
    }
    return total * sign;
  }
}

What to improve

  • ability to create my own edge cases.