43. Multiply Strings
题目
Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.
Note:
The length of both num1 and num2 is < 110.
Both num1 and num2 contains only digits 0-9.
Both num1 and num2 does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.
解析
- 大数相乘原则
- 认清楚乘法从后面开始乘，vec[size-1]是最小位
// add 43. Multiply Strings
class Solution_43 {
public:
string multiply(string num1, string num2) {
int len1 = num1.size();
int len2 = num2.size();
vector<int> vec(len1+len2,0); // 初始化内存空间
//vec.reserve(len1 + len2);
for (int i = 0; i < len1; i++)
{
int k = i;
for (int j = 0; j < len2;j++)
{
//vec.push_back(a*b);
vec[k] += (num1[len1-1-i] - '0')*(num2[len2-1-j]-'0'); ////Calculate from rightmost to left
k++;
}
}
string ret="";
for (int i = 0; i < vec.size();i++)
{
if (vec[i]>=10)
{
vec[i+1] += vec[i] / 10;
vec[i] = vec[i] % 10;
}
//char temp[5];
//_itoa(vec[i], temp, 10);
//ret += temp;
char temp = vec[i] + '0';
ret += temp; //反着取得
}
//reserve(ret.begin(),ret.end());
//reverse(vec.begin(), vec.end()); //string没有反转函数，vector有
//判断第一个非0位 //size_t startpos = sum.find_first_not_of("0");
int flag = 0;
for (int i = ret.size()-1; i >=0;i--)
{
if (ret[i]!='0')
{
flag = i;
break;
}
}
int begin = 0, end = flag;
while (begin < end)
{
swap(ret[begin], ret[end]);
begin++;
end--;
}
return ret.substr(0,flag+1);
}
};
题目来源
