8.String to Integer (atoi)

Tags: Medium Math String

Links: https://leetcode.com/problems/string-to-integer-atoi/

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Implement atoi which converts a string to an integer.

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.

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: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) 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.

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class Solution {
public:
    int myAtoi(string s) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        removeBlank(s);
        int res = 0;
        //处理带负号的情况
        bool isNegative = false;
        if (s[0] == '-') {
            isNegative = true;
            s = s.substr(1);
        }
        else if (s[0] == '+') s = s.substr(1);

        //处理第一个字符不是数字的情况
        if (!isdigit(s[0])) return res;

        int pos = 0, n = s.size();
        while (pos < n && isdigit(s[pos])) ++pos;
        s = s.substr(0, pos); //此时的S只包含数字

        pos = 0;
        while (s[pos] == '0') pos++;
        s = s.substr(pos);

        //处理越界的情况
        string MIN = to_string(INT_MIN);
        MIN = MIN.substr(1);
        if (isNegative && s.size() >= MIN.size()) {
            if (s.size() > MIN.size()) return INT_MIN;
            else if (s >= MIN) return INT_MIN;
        }
        else if (s.size() >= MIN.size()) {
            if (s.size() > MIN.size()) return INT_MAX;
            else if (s >= to_string(INT_MAX)) return INT_MAX;
        }

        for (int i = 0; i < s.size(); ++i) {
            res = res * 10 + (s[i] - '0');
        }
        if (isNegative) res *= -1;

        return res;
    }

    void removeBlank(string &s)
    {
        int pos = 0;
        while (s[pos] == ' ') ++pos;
        s = s.substr(pos);
    }
};

最初的写法代码很长，也不是很优雅，所以进行精简。

class Solution {
public:
    int myAtoi(string str) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (str.empty()) return 0;
        int sign = 1, base = 0, i = 0, n = str.size();
        while (i < n && str[i] == ' ') ++i;
        if (i < n && (str[i] == '+' || str[i] == '-')) {
            sign = (str[i++] == '+') ? 1 : -1;
        }
        //处理越界的情况
        while (i < n && str[i] >= '0' && str[i] <= '9') {
            if (base > INT_MAX / 10 || (base == INT_MAX / 10 && str[i] - '0' > 7)) {
                return (sign == 1) ? INT_MAX : INT_MIN;
            }
            base = 10 * base + (str[i++] - '0');
        }
        return base * sign;
    }
};

当然了，还有一种投机取巧的方法，可以速度最快，但是面试的时候肯定要GG的那种：

class Solution {
public:
    int myAtoi(string str) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        long long val = atoll(str.c_str());
        if (val > 2147483647)
            return 2147483647;
        if (val < -2147483647)
            return -2147483648;
        return (int)val;
    }
};