LeetCodeOJ. String to Integer (atoi)

试题请參见: https://oj.leetcode.com/problems/string-to-integer-atoi/

题目概述

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.

解题思路

这道题目还真是艰辛~ 看起来非常基础, 但是有些Case还真是暗藏玄机.

主要问题还是推断是否越界. 

我的做法是, 记录n = n * 10 + currentDigit运算前n的值. 若运算后n / 10和之前记录下来的值相等, 则还未越界.

源码

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');
    }
};

转载于:https://www.cnblogs.com/blfshiye/p/5114200.html