LeetCode 8. String conversion integer (ATOI)

Please come to realize a myAtoi(string s) function , Enable it to convert a string to a 32 Bit signed integer （ similar C/C++ Medium atoi function ）.

function myAtoi(string s) The algorithm is as follows ：

Read in strings and discard useless leading spaces
Check the next character （ Suppose you haven't reached the end of the character yet ） Positive or negative , Read the character （ If there is ）. Determine whether the final result is negative or positive . If neither exists , Suppose the result is positive .
Read in the next character , Until you reach the next non numeric character or the end of the input . The rest of the string will be ignored .
Convert the numbers read in the previous steps into integers （ namely ,“123” -> 123, “0032” -> 32）. If you don't read in the numbers , Then the integer is 0 . Change the symbol if necessary （ From step 2 Start ）.
If the number of integers exceeds 32 Bit signed integer range [−2³¹, 2³¹ − 1] , You need to truncate this integer , Keep it in this range . say concretely , Less than −2³¹ The integer of should be fixed to −2³¹ , Greater than 2³¹ − 1 The integer of should be fixed to 2³¹ − 1 .
Returns an integer as the final result .
Be careful ：

The white space character in this question only includes the space character ’ ’ .
Except for the leading space or the rest of the string after the number , Do not ignore Any other character .

Example 1：
Input ：s = “42”
Output ：42
explain ： The bold string is the character that has been read in , The caret is the character currently read .
The first 1 Step ：“42”（ No characters are currently read in , Because there are no leading spaces ）
^
The first 2 Step ：“42”（ No characters are currently read in , Because it doesn't exist here ‘-’ perhaps ‘+’）
^
The first 3 Step ：“42”（ Read in “42”）
^
Parse to get an integer 42 .
because “42” In scope [-231, 231 - 1] Inside , The final result is 42 .

Example 2：

Input ：s = " -42"
Output ：-42
explain ：
The first 1 Step ：" -42"（ Read in leading space , But ignore ）
^
The first 2 Step ：" -42"（ Read in ‘-’ character , So the result should be negative ）
^
The first 3 Step ：" -42"（ Read in “42”）
^
Parse to get an integer -42 .
because “-42” In scope [-231, 231 - 1] Inside , The final result is -42 .

Example 3：
Input ：s = “4193 with words”
Output ：4193
explain ：
The first 1 Step ：“4193 with words”（ No characters are currently read in , Because there are no leading spaces ）
^
The first 2 Step ：“4193 with words”（ No characters are currently read in , Because it doesn't exist here ‘-’ perhaps ‘+’）
^
The first 3 Step ：“4193 with words”（ Read in “4193”; Because the next character is not a number , So read in stop ）
^
Parse to get an integer 4193 .
because “4193” In scope [-231, 231 - 1] Inside , The final result is 4193 .

Code:

class Solution {
    
public:
    int myAtoi(string s) {
    
        int re = 0;
        int k = 0;
        while(k < s.size() && s[k] == ' ') k ++;
        if( k == s.size()) return 0;

        int flag = 1;
        if(s[k] == '-') flag = -1, k ++;
        if(s[k] == '+') 
        {
    
            if (flag == -1) return 0;
            else
                k ++;
        }
        while(k < s.size() && s[k] >='0' && s[k] <= '9')
        {
    
            int x = s[k] - '0';
            if(flag > 0 && re > (INT_MAX - x)/ 10) return INT_MAX;
            if(flag < 0 && -re < (INT_MIN + x)/ 10) return INT_MIN;
            if(-re * 10 - x == INT_MIN) return INT_MIN;   //  This sentence is a special judgment , INT_MN  and INT_MAX The absolute value of is different 
            re = re * 10 + x;
            k++;
        }
        re *= flag;
        return re;
    }
};