题目：
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.
spoilers alert... click to show 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.
题解：
这题主要就是考虑一下corner case。
越界问题？
正负号问题？
空格问题？
精度问题？
代码如下：
1 public int atoi(String str) {
2 if (str == null || str.length() < 1)
3 return 0;
4
5 // trim white spaces
6 str = str.trim();
7
8 char flag = '+';
9
// check negative or positive
int i = 0;
if (str.charAt(0) == '-') {
flag = '-';
i++;
} else if (str.charAt(0) == '+') {
i++;
}
// use double to store result
double result = 0;
// calculate value
while (str.length() > i && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
result = result * 10 + (str.charAt(i) - '0');
i++;
}
if (flag == '-')
result = -result;
// handle max and min
if (result > Integer.MAX_VALUE)
return Integer.MAX_VALUE;
if (result < Integer.MIN_VALUE)
return Integer.MIN_VALUE;
return (int) result;
}
更新：
因为遇见过了atol，string to long 这样的问题，就不能用这种比当前数据类型长的方法解决。
另一种方法是更加普遍和巧妙的，就是用这样的判断：
1. Integer.MAX_VALUE/10 < result; //当前转换结果比Integer中最大值/10还大（因为这个判断放在while循环最开始，之后还要对result进行*10+当前遍历元素的操作，所以如果还乘10的result已经比Integer.MAX_VALUE/10还大，可想而知，乘了10更大）
2. Integer.MAX_VALUE/10 == result && Integer.MAX_VALUE%10 <(str.charAt(i) - '0') //另外的情况就是，当前result恰跟 Integer.MAX_VALUE/10相等，那么就判断当前遍历的元素值跟最大值的最后一位的大小关系即可
代码如下：
1 public int atoi(String str) {
2 if (str == null || str.length() < 1)
3 return 0;
4
5 // trim white spaces at beginning and end
6 str = str.trim();
7
8 char flag = '+';
9
// check negative or positive
int i = 0;
if (str.charAt(0) == '-') {
flag = '-';
i++;
} else if (str.charAt(0) == '+') {
i++;
}
// use double to store result
int result = 0;
// calculate value
while (str.length() > i && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
if(Integer.MAX_VALUE/10 < result || (Integer.MAX_VALUE/10 == result && Integer.MAX_VALUE%10 < (str.charAt(i) - '0')))
return flag == '-' ? Integer.MIN_VALUE : Integer.MAX_VALUE;
result = result * 10 + (str.charAt(i) - '0');
i++;
}
if (flag == '-')
result = -result;
return result;
}
Reference： http://www.programcreek.com/2012/12/leetcode-string-to-integer-atoi/