String to Integer
最后更新于：2022-04-02 01:14:09
# String to Integer
### Source
- leetcode: [String to Integer (atoi) | LeetCode OJ](https://leetcode.com/problems/string-to-integer-atoi/)
- lintcode: [(54) String to Integer(atoi)](http://www.lintcode.com/en/problem/string-to-integeratoi/)
~~~
Implement function atoi to convert a string to an integer.
If no valid conversion could be performed, a zero value is returned.
If the correct value is out of the range of representable values,
INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
Example
"10" => 10
"-1" => -1
"123123123123123" => 2147483647
"1.0" => 1
~~~
### 题解
经典的字符串转整数题，边界条件比较多，比如是否需要考虑小数点，空白及非法字符的处理，正负号的处理，科学计数法等。最先处理的是空白字符，然后是正负号，接下来只要出现非法字符(包含正负号，小数点等，无需对这两类单独处理)即退出，否则按照正负号的整数进位加法处理。
### Java
~~~
public class Solution {
/**
* @param str: A string
* @return An integer
*/
public int atoi(String str) {
if (str == null || str.length() == 0) return 0;
// trim left and right spaces
String strTrim = str.trim();
int len = strTrim.length();
// sign symbol for positive and negative
int sign = 1;
// index for iteration
int i = 0;
if (strTrim.charAt(i) == '+') {
i++;
} else if (strTrim.charAt(i) == '-') {
sign = -1;
i++;
}
// store the result as long to avoid overflow
long result = 0;
while (i < len) {
if (strTrim.charAt(i) < '0' || strTrim.charAt(i) > '9') {
break;
}
result = 10 * result + sign * (strTrim.charAt(i) - '0');
// overflow
if (result > Integer.MAX_VALUE) {
return Integer.MAX_VALUE;
} else if (result < Integer.MIN_VALUE) {
return Integer.MIN_VALUE;
}
i++;
}
return (int)result;
}
}
~~~
### 源码分析
符号位使用整型表示，便于后期相乘相加。在 while 循环中需要注意判断是否已经溢出，如果放在 while 循环外面则有可能超过 long 型范围。
### 复杂度分析
略
### Reference
- [String to Integer (atoi) 参考程序 Java/C++/Python](http://www.jiuzhang.com/solutions/string-to-integer-atoi/)
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