Given two numbers represented as strings, return multiplication of the numbers as a string.
Note: The numbers can be arbitrarily large and are non-negative.
模拟法
根据乘法规则,乘数逐位与被乘数相乘,然后相加,因此先实现乘数只有一位的情况:
为了防止进位处理,把结果reverse保存,即高位在后面,地位在前面,然后被乘数和乘数逐一相乘,如果大于10,则需要进位:
string multiply(string num1, int i) {
if (i == 0)
return "0";
if (i == 1) {
reverse(begin(num1), end(num1)); // 返回结果总是reverse的
return num1;
}
string result;
int n = num1.size();
int j = n - 1;
int carry = 0;
for (int j = n - 1; j >= 0; --j) {
int c = (num1[j] - '0') * i + carry;
carry = c / 10;
c %= 10;
result.push_back(c + '0');
}
if (carry != 0)
result.push_back(carry + '0');
//reverse(begin(result), end(result));
return result;
}
算得每个结果后,需要把所有的相乘结果相加,因此需要实现大数加法:
/**
注意num1 和 num2都是reverse的,并且结果也是reverse的
*/
string add(string num1, string num2) {
string result;
int i = 0, j = 0;
int carry = 0;
while (i < num1.size() && j < num1.size()) {
int a = num1[i] - '0';
int b = num2[j] - '0';
int c = a + b + carry;
carry = c / 10;
c %= 10;
result.push_back(c + '0');
i++, j++;
}
while (i < num1.size()) {
int c = num1[i] - '0' + carry;
carry = c / 10;
c %= 10;
result.push_back(c + '0');
i++;
}
while (j < num2.size()) {
int c = num2[j] - '0' + carry;
carry = c / 10;
c %= 10;
result.push_back(c + '0');
j++;
}
if (carry != 0) {
result.push_back(carry + '0');
}
return result;
}
最后结果就是乘数逐位与被乘数相乘即可,注意乘数不是各位时,应该乘以10^n,n位第几位:
string multiply(string num1, string num2) {
if ( num1 == "0" || num1 == ""
|| num2 == "0" || num2 == "") {
return "0";
}
/*
* 反而更慢?
if (num1.size() < num2.size())
return multiply(num2, num1);
*/
string result;
int n = num2.size();
for (int i = n - 1; i >= 0; --i) {
if (num2[i] == '0')
continue;
string s = multiply(num1, num2[i] - '0');
string zeros;
for (int j = 1; j < n - i; ++j)
zeros.push_back('0');
s = zeros + s;
result = add(result, s);
}
reverse(begin(result), end(result));
return result == "" ? "0" : result;
}