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 };