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.
class Solution {
public:
string multiply(string num1, string num2) {
}
};
我的解法是每次提取num2的一位，然后和num1相乘，所得结果并入string res中，这样虽然每次都要新定义一个string，效率略低，但是思路比较清晰：把字符串相乘划分成了“字符串和数字相乘” 和 “字符串相加” 两个子问题，分别解决就可以。
处理时先将num1，和num2逆序，使得最低位在[0] 位置，这样处理起来比较方便，不易写错。
class Solution {
public:
string multiply(string num1, string num2) {
if(num1.length() == || num2.length() == )
return "";
if(num1.length() == && num1[] == '') return "";
if(num2.length() == && num2[] == '') return "";
string res(num1.length() + num2.length(), '');
int i = , j = ;
reverse(num1.begin(), num1.end());
reverse(num2.begin(), num2.end());
for(i = ; i < num2.length(); ++i){
int offset = i;
string tmp(offset + num1.length() + , '');
int add = ;
for(j = ; j < num1.length(); ++j){
int tmpMul = (num1[j] - '') * (num2[i] - '') + add;
add = tmpMul / ;
tmp[j + offset] = (tmpMul % ) + '';
}
tmp[j + offset] = add + '';
plus(res, tmp);
}
int countStartZero = ; //高位'0'的个数
for(i = res.length() - ; i >= && res[i] == ''; --i, ++countStartZero);
if(countStartZero == res.length()) return "";
res = res.substr(, res.length() - countStartZero);
reverse(res.begin(), res.end());
return res;
}
private:
void plus(string &s1, string &s2){
int add = ;
for(int i = ; i < s1.length() && (add > || i < s2.length()); ++i){
int tmp = (s1[i] - '') + add;
if(i < s2.length()) tmp += (s2[i] - '');
add = tmp / ;
s1[i] = (tmp % ) + '';
}
}
};