題目：

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.

例子:

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.

問題解析：

實現 atoi，將字串轉為整數。

思路標籤

邊界檢查

解答：

直接轉換

• 問題簡單，最主要的是需要注意邊界條件；
• 首先解決前置空格；
• 其次判斷正負號；
• 再次接下來一個非空字元不為數字，則整體不符合轉換要求；
• 接下來遇到非數字字元，對字串進行截斷；
• 通過long型儲存和正負號flag來進行判斷是否越界。

class Solution {
public:
    int myAtoi(string str) {
        if(str.empty())
            return 0;
        long digit = 0;
        int minusflag = 1;
        int i = 0;
        while(str[i] == ' ')
            i++;
        if(str[i] == '-'){
            minusflag = -1;
            i++;
        }
        else if(str[i] == '+')
            i++;

        if(str[i] - '0' < 0 || str[i] - '0' > 9)
            return 0;

        while(str[i] - '0' >= 0 && str[i] - '0' <= 9 && digit < INT_MAX && i < str.size()){
            int num = str[i] - '0';
            digit = digit*10 + num;
            i++;
        }
        if(digit*minusflag > INT_MAX)
            return INT_MAX;
        if(digit*minusflag < INT_MIN)
            return INT_MIN;
        return digit*minusflag;
    }
};