Leetcode 8. String to Integer (atoi)

题目

Implement atoi which converts a string to an integer.

The function first discards as many whitespace characters as neccesssary 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: [-2^31, 2^32 - 1]. If the numerical value is out of the range of repressentable values, INT_MAX (2^31 - 1) or INT_MIN (-2^31) is returned.

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.

Example 5:

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.

Thoughts

The most challenging part of this question is to handle all tricky edge cases. Personally I would approach this problem with the following three steps: 1. Skip all whitespace in the beginning (also handle the whitespace only case). 2. Check sign if the first non whitespace character is a sign. 3. Keep constructing result until first non-digit is met (also handle the overflow case).

Solution

The following solution strictly follows above three steps for better readability. Yet, it uses two while loops which are unneccessary. Although personally I prefer the current solution that are easier to follow through (and debug), we can opt to use one single loop and check all possible cases within it. By this means, the algorithm is generally faster, despite they have the same time complexity.

class Solution {
    public int myAtoi(String str) {

        int i = 0;
        int sign = 1;
        long res = 0;

        // skip white space
        while (i < str.length() && str.charAt(i) == ' ') {
            i++;
        }

        // all white space
        if (i == str.length()) return 0;

        // check sign
        char first = str.charAt(i);
        if (first == '+' || first == '-') {
            i++;
            if (first == '-') sign = -1;
        }

        // construct the result while digits
        while (i < str.length() && Character.isDigit(str.charAt(i))) {
            res = res * 10 + Character.getNumericValue(str.charAt(i));
            i++;
            // check boundary
            if (res * sign > Integer.MAX_VALUE) return Integer.MAX_VALUE * sign;
            if (res * sign < Integer.MIN_VALUE) return Integer.MIN_VALUE * sign;
        }

        return (int) res * sign;
    }
}

Complexity Analysis

• Time Complexity: O(n). We have to iterate through all characters in the string.

• Space Complexity: O(1). We don't use any extra space.

Extension

If we are asked to convert string's all digital characters to integer (instead of the first sequence of digital chars), are you able to do it?

Summary

Edge cases are important. We have to think thoroughly to cover all cases in a timely manner and then find solutions that handle them with minimum conditional checks. An elegant solution is the one that cover many complicated cases with concise codes.

I realize on Leetcode, the questions that require deliberate consideration on edge cases recieve most downvotes. However, we do need to practice dealing with complex edge cases.

Reference

• Mao's original series