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multiply-strings.cpp
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59 lines (51 loc) · 1.76 KB
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class Solution {
public:
//比较麻烦,也不能算是自己做的,主要参考了这个:https://www.cnblogs.com/springfor/p/3889706.html
string multiply1(string num1, string num2) {
reverse(num1.begin(),num1.end());
reverse(num2.begin(),num2.end());
int length1=num1.size(),length2=num2.size();
int carry=0;
string result="";
vector<int> r(length1+length2,0);
for(int i=0;i<length1;i++){
for(int j=0;j<length2;j++){
r[i+j]+=(num1[i]-'0')*(num2[j]-'0');
cout<<r[i+j]<<endl;
}
}
for(int i=0;i<length1+length2;i++){
int digit=r[i]%10;
carry=r[i]/10;
result.insert(result.begin()+0,digit+'0');
if(i<length1+length2-1) r[i+1]+=carry;
}
cout<<result<<endl;
for(int i=0;i<length1+length2;i++){
if(result[i]!='0'){
// cout<<"i:"<<i<<":"<<result[i]<<endl;
result=result.substr(i,length1+length2-i);
break;
}
if(i==length1+length2-1) return "0";
}
return result;
}
string multiply(string num1, string num2) {
string sum(num1.size() + num2.size(), '0');
for (int i = num1.size() - 1; 0 <= i; --i) {
int carry = 0;
for (int j = num2.size() - 1; 0 <= j; --j) {
int tmp = (sum[i + j + 1] - '0') + (num1[i] - '0') * (num2[j] - '0') + carry;
sum[i + j + 1] = tmp % 10 + '0';
carry = tmp / 10;
}
sum[i] += carry;
}
size_t startpos = sum.find_first_not_of("0");
if (string::npos != startpos) {
return sum.substr(startpos);
}
return "0";
}
};