【剑指offer】把字符串转换成整数(溢出判断)
把字符串转换成整数
https://www.nowcoder.com/practice/1277c681251b4372bdef344468e4f26e?tpId=13&tqId=11202&tPage=3&rp=3&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
class Solution {
public:
int StrToInt(string str) {
//关键是判断整数的上下界溢出,利用INT_MIN INT_MAX
int n = str.size(), s = 1;
long long res = 0;
if(!n) return 0;
if(str[0] == '-') s = -1;
for(int i = (str[0] == '-' || str[0] == '+') ? 1 : 0; i < n; ++i){
if(!('0' <= str[i] && str[i] <= '9')) return 0;
if(s == 1 && (res > INT_MAX / 10 || res == INT_MAX / 10 && (str[i] - '0') > INT_MAX % 10)){
res = 0;
break;
}
if(s == -1 && (res < INT_MIN / 10 || res == INT_MIN / 10 && (str[i] - '0') > -(INT_MIN % 10))){
res = 0;
break;
}//边界条件值得记住
res = (res << 1) + (res << 3) + s*(str[i] & 0xf);//res=res*10+str[i]-'0';
}
return res;
}
};