8.String to Integer

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题目描述

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.

我的解法

解题思路

将一个字符串转化成数组，首先遇到空格要跳过，然后遇到的第一个+号和-号就是数字的符号。然后接下来的数字就是我们要的数字，后面有字符也可以，但是如果字符在数字前面先出现就不可以。并且要做int型的溢出检测，在之前一道题就遇到了https://yutouwd.github.io/posts/837075，就是下一题哈哈哈哈哈。

实现代码

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class Solution {
public:
    int myAtoi(string str) {
        int i = 0, n = str.size();
        bool flag = true;
        int ans = 0;
        if (n == 0)
            return 0;
        while (str[i] == ' ')
            i++;
        if (str[i] == '-' || str[i] == '+'){
            if (str[i] == '-')
                flag = false;
            i++;
        }
        while ((i < n) && isdigit(str[i])){
            int num = str[i] - '0';
            if (flag && ((ans > INT_MAX/10) || (ans == INT_MAX/10 && num > 7))) return INT_MAX;
            if (!flag && ((ans > -INT_MIN/10) || (ans == -INT_MIN/10 && num >= 8))) return INT_MIN;
            ans = ans * 10 + num;
            i++;
        }
        if (flag)
            return ans;
        else
            return -ans;
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for String to Integer (atoi).
Memory Usage: 8.4 MB, less than 91.04% of C++ online submissions for String to Integer (atoi).