String to Integer (atoi)

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... 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.

 

该题最好先考虑各种输入情况，再动手写代码

输入：

空串

125

12.36

546a

#4655

     45656

-0001

2147483647

-2147483648

9999999999999999

-99999999999999

 

class Solution {
public:
    int atoi(const char *str) {
        if( !str ) return 0;
        while( *str && *str == ' ' ) ++str; //寻找第一个非空格字符
        bool minus = false;
        if( *str == '-' ) { //判断是否为负
            minus = true;
            ++str;
        }
        else if( *str == '+' ) ++str;   //或'+'
        long long ans = 0;  //为防止溢出，使用long long
        while( *str && isdigit(*str) ) {    //碰到非数字或结束符就结束
            ans = ans * 10 + (*str - '0');
            if (minus && -ans < INT_MIN) {  // 处理下溢
                return INT_MIN;
            }
            if (!minus && ans > INT_MAX) {  // 处理上溢
                return INT_MAX;
            }

            ++str;
        }

        return minus ? -ans : ans;   //还原真实值
    }
};

 

class Solution {
public:
    int myAtoi(string str) {
        int index = 0;
        for (; index < str.length(); ++index) {
            if (str[index] != ' ') {
                break;
            }
        }

        if (index >= str.length()) {
            return 0;
        }

        bool minus;
        if (str[index] == '+') {
            minus = false;
            ++index;
        } else if (str[index] == '-') {
            minus = true;
            ++index;
        } else {
            minus = false;
        }

        long long ans = 0;
        while (index < str.length() && isdigit(str[index])) {
            ans = ans * 10 + (str[index] - '0');
            if (minus && -ans < INT_MIN) {
                return INT_MIN;
            }
            if (!minus && ans > INT_MAX) {
                return INT_MAX;
            }
            ++index;
        }

        return minus ? -ans : ans;
    }
};