LeetCodeOJ. String to Integer (atoi)

試題請參見: https://oj.leetcode.com/problems/string-to-integer-atoi/

題目概述

Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
spoilers alert...
Requirements for atoi:
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. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

解題思路

這道題目還真是艱辛~ 看起來很基礎, 可是有些Case還真是暗藏玄機.

主要問題還是判斷是否越界. 

我的做法是, 記錄n = n * 10 + currentDigit運算前n的值. 若運算後n / 10和之前記錄下來的值相等, 則還未越界.

源代碼

class Solution {
public:
    int atoi(const char *str) {
        int n = 0;
        bool isPositive = true;
        size_t i = 0;
        
        // Ignore Spaces
        for ( ; str[i] == ' ' && str[i] != 0; ++ i ) { }
        
        // Process Sign Bit
        if ( str[i] == '+' || str[i] == '-' ) {
            isPositive = (str[i] == '+');
            ++ i;
        }

        // Convert to Integer
        for ( ; isDigit(str[i]) && str[i] != 0; ++ i ) {
            char digit = str[i] - '0';
            int previousResult = n;
            n = n * 10 + digit;

            // If it's Overflow
            if ( n / 10 != previousResult ) {
                if ( isPositive ) {
                    return INT_MAX;
                } else {
                    return INT_MIN;
                }
            }
        }
        
        return ( isPositive ? n : -n );
    }
private:
    bool isDigit(char digit) {
        return (digit >= '0' && digit <= '9');
    }
};