//
// Created by HINTS on 2018/11/29.
//
#include <iostream>
#include <string>
using namespace std;
int myAtoi(string str){
int i = 0;
int base = 0;
while (isspace(str[i])){
i++;
}
int sign = 1;
if(str[i]=='+' || str[i] =='-'){
sign = (str[i] == '-')?-1:1;
i++;
}
while(isdigit(str[i])){
int digit = str[i] - '0';
//发生溢出
if(sign == -1){
if(-base < (INT_MIN + digit)/10)
return INT_MIN;
} else{
if (base > (INT_MAX - digit)/10)
return INT_MAX;
}
base = base*10 + digit;
i++;
}
return base*sign;
}
int main() {
cout<<myAtoi("2147483648")<<endl;
return 0;
}
心得:
关键是控制溢出,INT_MAX和INT_MIN,网上有一些答案,但运行结果不对。最后试了这个,运行通过。