# **8-String to Integer(Atoi)**
###### tags: `Medium`
## Question
https://leetcode.com/problems/string-to-integer-atoi/

- 情況大致分成有空格或詞彙在數字前、數字前有兩個以上正負號、和數字後有空格和詞彙、小數型式，其實只要根據規則去寫就能解開
## Solution
1. 空格要忽略
2. 正負號要檢查，但正負號或空格相連都回傳0
3. 數字出現就代表開始，中間有標點符號就是結束，不論後面有無數字再出現
4. 數值一定要在INT_MAX~INT_MIN之間
### C++

class Solution {
public:
    int myAtoi(string s) {
       long long int num = 0;
        int n =0;
        int start = 0;
        for(int i = 0; i < s.size();i
++){
            cout<<s[i];
            if(s[i] == ' ')
            {
                if(start == 1)
                {
                    if(n == 1)
                    {
                        return (-1)*num;
                    }
                    return num;
                }
                continue;   
            }
            if(s[i] == '-')
            {
                if(start == 1)
                {
                    if(n == 1)
                    {
                        return (-1)*num;
                    }
                    return num;
                }
                if(s[i+1] > '9' || s[i+1] < '0')
                {
                    return 0;
                }
                n = 1;
            }
            else if(s[i] == '+')
            {
                if(start == 1)
                {
                    if(n == 1)
                    {
                        return (-1)*num;
                    }
                    return num;
                }
                if(s[i+1] > '9' || s[i+1] < '0')
                {
                    return 0;
                }
                n = 0;
            }
            else if(s[i] <= '9' && s[i] >= '0')
            {
                start = 1;
                num = s[i]-'0' + num*10;  
                if(num > pow(2,31)-1)
                {
                    if(n == 1)
                    {
                        return (-1)*pow(2,31);
                    }
                    else
                    {
                        return pow(2,31)-1;
                    }
                }
            }
            else
            {
                if(start == 1)
                {
                    if(n == 1)
                    {
                        return (-1)*num;
                    }
                    return num;
                }
                break;
            }
            
        }
        if(n == 1)
        {
            return num*(-1);
        }   

        return num;
    }
};

### Python
- 正則表達式

class Solution:
    def myAtoi(self, s):
        m=re.match(r'\s*([\+\-]?\d+)',s)
        if not m: return 0
        
        ans=int(m.group(1))
        
        if ans>2**31-1:
            ans=2**31-1
        elif ans<-2**31:
            ans=-2**31
        
        return ans

## Solution
- 大神想法:

class Solution(object):
    def myAtoi(self, s):
        """
        :type s: str
        :rtype: int
        """
        digits = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
        signs = ['+', '-']
        negative = False
        numeric = False
        value = 0
        maxInt = 2**31-1
        minInt = -2**31
        
        # return for empty strings
        if len(s) == 0:
            return 0
        
        # skip whitespace
        index = 0
        while index < len(s) and s[index] == ' ':
            index += 1
        
        # determine the sign
        if index < len(s) and s[index] in signs:
            if s[index] == '-':
                negative = True
            index += 1
        
        # find the numeric value
        trimmed = ""
        while index < len(s) and (s[index] in digits):
            if not numeric:
                value = int(s[index])
                numeric = True
            else:
                value = value*10 + int(s[index])
            index += 1
            
        if not numeric:
            return 0
        if negative:
            value = value * -1
        if value < minInt:
            return minInt
        if value > maxInt:
            return maxInt
        return value

## Solution
- programming in C++

    int n = s.size();
    int num = 0;
    int flg = 0;
    int i = 0;
  
    while(i<n && s[i] == ' '){
        i++;
    }
    
    
    
    if(i<n && (s[i] == '+' || s[i] == '-')){
        if(s[i] == '-') flg = 1; 
            i++;
    }
    
    if( i<n && s[i] < '0' || s[i] > '9') return 0;
    
    while(i<n && s[i] == '0'){
        i++;
    }
    
    if(i<n && (s[i] < '0' || s[i] > '9')) return 0;
    
    while(i<n && s[i] >= '0' && s[i] <= '9'){
        if(num > INT_MAX/10){
            if(flg == 1) return INT_MIN;
            return INT_MAX;
        }
        if(num == INT_MAX/10 && (s[i]-'0') > 7){
             if(flg == 1) return INT_MIN;
            return INT_MAX;
        } 
        num = num*10 + (s[i]-'0');
        i++;
    }
    
    if(flg == 1) num *=-1;
    
    return num;
}