题目描述
Implement atoi which converts a string to an integer.
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.
NOTE
- Only the space character ‘ ‘ is considered as whitespace character.
- Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.
Example1:
Input: " -42" Output: -42 Explanation: The first non-whitespace character is '-', which is the minus sign.Then take as many numerical digits as possible, which gets 42.
Example2:
Input: "words and 987" Output: 0 Explanation: The first non-whitespace character is 'w', which is not a numerical digit or a +/- sign. Therefore no valid conversion could be performed.
Example3:
Input: "-91283472332" Output: -2147483648 Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.Thefore INT_MIN (−2^31) is returned.
解题思路
- 根据题意，首先判断是不是连续的空格
- 然后判断正负号
- 最后在生成最后的value时，看是否溢出了（跟INT_MAX进行对比，如果溢出了，并且是负数，返回 INT_MIN,否则返回 INT_MAX）
class Solution {
public:
int myAtoi(string str) {
if(str.size() == 0){
return 0;
}
int i = 0;
int sign = 1;
unsigned long value = 0;
while(i < str.size() && str[i] == ' '){
i++;
}
if(str[i] == '-' || str[i] == '+'){
if(str[i] == '-'){
sign = -1;
}
i++;
}
for(; i < str.size() && str[i] >= '0' && str[i] <= '9'; i++){
value = value * 10 + (str[i] - '0');
if(value > INT_MAX){
return sign == -1 ? INT_MIN : INT_MAX;
}
}
return value * sign;
}
};
