题目来源：https://leetcode.com/problems/string-to-integer-atoi/
Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.
解题思路：
照着要求写代码，可以总结如下：
1. 字串为空或者全是空格，返回0；
2. 字串的前缀空格需要忽略掉；
3. 忽略掉前缀空格后，遇到的第一个字符，如果是‘+’或‘－’号，继续往后读；如果是数字，则开始处理数字；如果不是前面的2种，返回0；
4. 处理数字的过程中，如果之后的字符非数字，就停止转换，返回当前值；
5. 在上述处理过程中，如果转换出的值超出了int型的范围，就返回int的最大值或最小值。
Java代码：
1 public class Solution { 2 public int myAtoi(String str) { 3 int max = Integer.MAX_VALUE; 4 int min = -Integer.MIN_VALUE; 5 long result = 0; 6 str = str.trim(); 7 int len = str.length(); 8 if (len < 1) 9 return 0; 10 int start = 0; 11 boolean neg = false; 12 13 if (str.charAt(start) == '-' || str.charAt(start) == '+') { 14 if (str.charAt(start) == '-') 15 neg = true; 16 start++; 17 } 18 19 for (int i = start; i < len; i++) { 20 char ch = str.charAt(i); 21 22 if (ch < '0' || ch > '9') 23 break; 24 result = 10 * result + (ch - '0'); 25 if (!neg && result > max) 26 return max; 27 if (neg && -result < min) 28 return min; 29 30 } 31 if (neg) 32 result = -result; 33 34 return (int) result; 35 } 36 37 };