leetcode——String to Integer (atoi) 字符串转换为整型数(AC)
2014-06-06 22:04
501 查看
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.
这道题的要求还是很多的,很多的case都需要注意,需要满足要求,例如开头的空格、开头的+、-符号标志、字符满足数字的范围,并且需要满足int型整数的正、负边界。关于符号位的判定,我用了一个bool变量来做标记,关于int型边界的判定,我采用的方法是在每次乘10做加法之前,先使用最大边界值做减法然后除以10判定是否会溢出,在不溢出的情况下再继续进行运算。代码如下:
class Solution {
public:
int atoi(const char *str)
{
bool minus = false;
int result = 0;
const char *temp = str;
unsigned int MAX = 2147483647;
unsigned int MIN = 2147483648;
while(*temp == ' ')
*temp++;
if(*temp == '-')
{
minus = true;
temp++;
}
else if(*temp == '+')
{
minus = false;
temp++;
}
if(*temp==NULL || *temp<'0' || *temp>'9')
return 0;
while(*temp >= '0' && *temp <= '9')
{
if(!minus && (MAX-(int)(*temp-'0'))/10 >= result)
{
result = result*10+(int)(*temp - '0');
}
else if(minus && (MIN-(int)(*temp-'0'))/10 >= result)
{
result = result*10+(int)(*temp - '0');
}
else
{
return minus ? INT_MIN : INT_MAX;
}
temp++;
}
if(minus)
result = 0-result;
return result;
}
};
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.
这道题的要求还是很多的,很多的case都需要注意,需要满足要求,例如开头的空格、开头的+、-符号标志、字符满足数字的范围,并且需要满足int型整数的正、负边界。关于符号位的判定,我用了一个bool变量来做标记,关于int型边界的判定,我采用的方法是在每次乘10做加法之前,先使用最大边界值做减法然后除以10判定是否会溢出,在不溢出的情况下再继续进行运算。代码如下:
class Solution {
public:
int atoi(const char *str)
{
bool minus = false;
int result = 0;
const char *temp = str;
unsigned int MAX = 2147483647;
unsigned int MIN = 2147483648;
while(*temp == ' ')
*temp++;
if(*temp == '-')
{
minus = true;
temp++;
}
else if(*temp == '+')
{
minus = false;
temp++;
}
if(*temp==NULL || *temp<'0' || *temp>'9')
return 0;
while(*temp >= '0' && *temp <= '9')
{
if(!minus && (MAX-(int)(*temp-'0'))/10 >= result)
{
result = result*10+(int)(*temp - '0');
}
else if(minus && (MIN-(int)(*temp-'0'))/10 >= result)
{
result = result*10+(int)(*temp - '0');
}
else
{
return minus ? INT_MIN : INT_MAX;
}
temp++;
}
if(minus)
result = 0-result;
return result;
}
};
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