LeetCode:String to Integer (atoi)

Posted 2020-06-20 翎飞蝶舞

8. String to Integer (atoi)

 

 

Total Accepted: 91403 Total Submissions: 683692 Difficulty: Easy

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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

spoilers alert... click to show requirements for atoi.

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.

 

    类似于上一题，同样定义res为long类型用于避免溢出的情况，先算出res的值再判断是否溢出。

class Solution {
public:
    int myAtoi(string str) {
        if (str.length() ==0)
            return 0;
        int i = 0;
        while (str[i] == ‘ ‘)
            i++;
        if (i == str.length())
            return 0;
        long long res=0;
        int isPo = 1;
        if (str[i] == ‘-‘)
        {
            isPo = -1;
            i++;
        }
        else if (str[i] == ‘+‘)
        {
            isPo = 1;
            i++;
        }
        for (i; i < str.length(); i++)
        {
            if (str[i]<‘0‘ || str[i]>‘9‘)
                break;
            res = res * 10 + str[i]-‘0‘;
        }
        res = res*isPo;
        if (res>INT_MAX)
            return INT_MAX;
        else if (res < INT_MIN)
            return INT_MIN;
        return res;
    }
};

        但上一题中原始用于变换的数即是int型，位数固定，翻转后的数值也不会超过long型的表示范围。而该题目中的字符串可能是无限长的，所以long型仍可能溢出。所以我们要再每次给res加进一位数时就判断一下是否溢出。

for (i; i < str.length(); i++)
        {
            if (str[i]<‘0‘ || str[i]>‘9‘)
                break;
            res = res * 10 + str[i] - ‘0‘;
            if (isPo*res>INT_MAX)
                return INT_MAX;
            else if (isPo*res < INT_MIN)
                return INT_MIN;
        }