8. 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.
题目大意：该题目是说将string类型的字符串转换成整型数据，类似于C++库里的atoi函数，解决该题目的关键在于两个方面：
（1）字符串格式的合法判断
（2）转换结果的溢出判断
首先，对于字符串格式，空格不计入计算，应从第一个非空字符开始判断，首字母只能是符号（+、-）与数字的一种；从计算开始遍历字符串，到最后一位数字为止；
其次，对于转换结果，我们知道整型数据的范围是INT_MIN（-2147482648）到INT_MAX（2147483647），超出范围则返回最大与最小值。所以我们可以开始用long long类型的变量存储结果；
代码如下：
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class
Solution {
public
:
int
myAtoi(string str) {
if
(str.empty())
return
0;
int
flag = 1;
//flag 1正 -1负
long
long
result = 0;
int
i = 0;
while
(str[i] ==
' '
)
{
i++;
}
if
(str[i] ==
'-'
)
{
i++;
flag = -1;
}
else
if
(str[i] ==
'+'
)
{
i++;
}
for
(
int
j = i; j < str.length(); j++)
{
if
(str[j] >=
'0'
&& str[j] <=
'9'
)
{
result = result * 10 + (str.at(j) -
'0'
);
if
(result > 2147483647)
{
if
(flag == 1)
result = INT_MAX;
else
{
result = INT_MIN;
flag = 1;
}
break
;
}
}
else
break
;
}
return
flag * result;
}
};
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本文转自313119992 51CTO博客，原文链接:http://blog.51cto.com/qiaopeng688/1835919
