# 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/)