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.
实现字符串转整数的函数，需要注意的几点： 本文地址
1、对于非法输入，函数输出0（除了溢出）
2、处理输入NULL
3、处理数字前面的空格
4、处理数字前面的0
5、处理溢出
6、注意“+ 123”输入是非法的
class Solution {
public:
int atoi(const char *str) {
if(str == NULL)return 0;//0、处理NULL
while(*str == ' ')str++;//1、过滤掉空格
bool sign = true;
if(*str == '-'){sign = false; str++;}//2、判断符号位
else if(*str == '+')str++;
while(*str == '0')str++;//3、过滤掉数字前面的0
long long res = 0, lastRes = 0;
while(*str != ' ' && *str >= '0' && *str <= '9')
{
lastRes = res;
res = res*10 + (int)(*str - '0');
if(res > INT_MAX)return sign == true ? INT_MAX : INT_MIN;//4、判断是否溢出
str++;
}
return sign == true ? res : res*-1;
}
};
上面是把res声明为longlong来判断溢出，亦可以如下操作
class Solution {
public:
int atoi(const char *str) {
if(str == NULL)return 0;//0、处理NULL
while(*str == ' ')str++;//1、过滤掉空格
bool sign = true;
if(*str == '-'){sign = false; str++;}//2、判断符号位
else if(*str == '+')str++;
while(*str == '0')str++;//3、过滤掉数字前面的0
int res = 0, lastRes = 0;
int limit = INT_MAX / 10;
while(*str != ' ' && *str >= '0' && *str <= '9')
{
if(limit < res)return sign == true ? INT_MAX : INT_MIN;//4、判断是否溢出(两个数相乘溢出后，溢出的结果不一定比两个乘数小)
lastRes = res;
res = res*10 + (int)(*str - '0');
if(res < lastRes)return sign == true ? INT_MAX : INT_MIN;//4、判断是否溢出
str++;
}
return sign == true ? res : res*-1;
}
};
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