面试题:
给定两个由0-9数字组成的最长可到30个字符的字符串,请计算他们对应的整数和。允许使用字符串
转最大不超过32bit整形的系统函数。
当我看到这个面试题的时候,貌似不是第一次,所以就动手写了写。欢迎在下面留言写其他方法。也可以加入QQ羣聊:83459374
好了不多说,请看代码:
void calculateAdd()
{
string str1 = “2142135213214235231323241”;
string str2 = “809327498327413628164321434214”;
string result = “”;
cout << “str1的长度为:“ << str1.length() <<endl;
cout << “str2的长度为:“ << str2.length() <<endl;
int quotient = 0;
int remainder = 0;
int count = 0;
int strNum1 = str1.length();
int strNum2 = str2.length();
int max = strNum1 > strNum2 ? strNum1: strNum2;
cout << “max的长度为:“ << max <<endl;
while (max > 0) {
strNum1–;
strNum2–;
max–;
int nStr1 = strNum1 >= 0 ? ((int)(str1.at(strNum1)) – 48):0;
int nStr2 = strNum2 >= 0 ? ((int)(str2.at(strNum2)) – 48):0;
count = nStr1 + nStr2 + quotient;
if (count > 9){
quotient = count / 10;
remainder = count % 10;
}else{
quotient = 0;
remainder = count;
}
result = (char)(remainder+48) + result;
}
if(quotient > 0){
result = (char)(quotient+48) + result;
}
cout << “result的长度为:“ << result.length() <<endl;
cout<<“———-和为:“<<result.c_str()<<endl;
}
输出结果为:
str1的长度为:25
str2的长度为:30
max的长度为:30
result的长度为:30
———-和为:809329640462626842399552757455
亲测了一下,目测应该是对的,没什么问题,欢迎更多交流!