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：
public class Solution {
public int myAtoi(String str) {
int max = Integer.MAX_VALUE;
int min = -Integer.MIN_VALUE;
long result = 0;
str = str.trim();
int len = str.length();
if (len < 1)
return 0;
int start = 0;
boolean neg = false;
if (str.charAt(start) == '-' || str.charAt(start) == '+') {
if (str.charAt(start) == '-')
neg = true;
start++;
}
for (int i = start; i < len; i++) {
char ch = str.charAt(i);
if (ch < '0' || ch > '9')
break;
result = 10 * result + (ch - '0');
if (!neg && result > max)
return max;
if (neg && -result < min)
return min;
}
if (neg)
result = -result;
return (int) result;
}
};