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# 8. String to Integer (atoi)

## 1.問題

## 2.想法

* 提問
  * 是不是只會輸出一個數字-> 不須轉換每個substring
  * 正負號
* function header, parameter
* test input
  * 空白 + 數字
  * integer overflow
* 說明想法
  * invalidate的情形: 
    * 字串 + 數字
    * 兩個正負號 + 數字
  * 忽略空白, 遇到正負號讀取後index + 1
  * 處理overflow
    * 加之前與加之後必須數值相同

```
backup = sum;
sum = sum * 10 + ans;
            
if (sum < 0 || backup != (sum - ans) / 10) {
   return sign > 0 ? INT_MAX : INT_MIN;
}
```

* 測試
* 計算複雜度: O(n)?

## 3.程式碼

```
class Solution {
public:
    int myAtoi(string str) {
        if (str.empty()) {
            return 0;
        }
        int size = str.length(), i = 0, sign = 1, sum = 0, backup = 0;
        
        //如果碰到空白就往後移
        while (i < size && str[i] == ' ') {
            i++;
        }
        
        //判斷符號
        if (str[i] == '+') {
            sign = 1;
            i++;
        } else if (str[i] == '-') {
            sign = -1;
            i++;
        }
        
        while (i < size && isdigit(str[i])) {
           
            int ans = int(str[i] - '0');
            backup = sum;
            sum = sum * 10 + ans;
            
            if (sum < 0 || backup != (sum - ans) / 10) {
                return sign > 0 ? INT_MAX : INT_MIN;
            }
            i++;
        }
        
        
        return sum * sign;
    }
};
```