[LeetCode] #43 Multiply Strings

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.

本题先用数组存放每一位相乘后的累加数字，比如几个十，然后把它转换为string类型的数字，再去掉前面多余的0.时间：9ms。代码如下：

class Solution {
public:
    string multiply(string num1, string num2) {
        if (num1.size() == 0)
            return num2;
        else if (num2.size() == 0)
            return num1;
        vector<int> v(num1.size() + num2.size() - 1, 0);
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());
        for (string::size_type i = 0; i < num2.size(); i++){
            if (num2[i] != '0'){
                for (string::size_type j = 0; j < num1.size(); j++){
                    v[i + j] += (num1[j] - '0')*(num2[i] - '0');
                }
            }
        }
        string str="";
        int temp(0);
        for (vector<int>::size_type i = 0; i < v.size(); i++){
            str += '0' + (v[i] + temp) % 10;
            temp = (v[i] + temp) / 10;
        }
        if (temp>0)
            str += '0' +  temp % 10;
        reverse(str.begin(), str.end());
        string::const_iterator iter = str.begin();
        for (; iter != str.end() && *iter == '0'; ++iter);
        if (iter == str.end())
            return "0";
        else
            return str.substr(iter - str.begin(), str.end() - iter);
    }
};