问题描述：
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.
Requirements for atoi:
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.
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.
解题思路：
首先调用trim方法除去字符串两边的空白字符。然后从开头读起，如果开头为‘+’，表明该值为正数；如果开头为‘-’，表明该值为负数；如果第一个读到的是‘+’或者‘-’，则从第二个开始读，否则仍然从第一个读起。下面开始一直往后读知道字符串结束。如果在读的过程中读到了除字符’0‘-’9‘以外的数字，那么跳出循环，返回计算好的值，否则，不停的计算累加值。
代码如下：
1 public class Solution {
2 public int myAtoi(String str) {
3 boolean negative = false;
4 int i = 0;
5 int result = 0;
6 int digit;
7
8 if (str == null || str.length() == 0)
9 return 0;
10 str = str.trim();
11 if (str.length() == 0)
12 return 0;
13 if (str.charAt(0) == ‘-‘ || str.charAt(0) == ‘+‘) {
14 if (str.charAt(0) == ‘-‘) {
15 negative = true;
16 }
17 i++;
18 }
19 while (i < str.length()) {
20 if (str.charAt(i) > ‘9‘ || str.charAt(i) < ‘0‘)
21 break;
22 digit = str.charAt(i) - ‘0‘;
23 if (!negative && result > (Integer.MAX_VALUE - digit) / 10)
24 return Integer.MAX_VALUE;
25 else if (negative && result > -((Integer.MIN_VALUE + digit) / 10))
26 return Integer.MIN_VALUE;
27 else {
28 result = result * 10 + digit;
29 i++;
30 }
31 }
32 return negative ? -result : result;
33 }
34 }