題幹：

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contain only digits 0-9.
3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.
4. You must not use any built-in BigInteger library or convert the inputs to integerdirectly.

大意：

　　大整數乘法

輸入：string

輸出：string

思路：

　　1、計算乘積中對應位的結果。不考慮進位時，根據乘數位數可以確定乘積位數，乘積對應位置的數是由乘數確定位置的數確定。如num1*num2，結果會有len1-1+len2-1+1位。計算時：

R[len - i - j]+=((num1[i]-‘0‘)*(num2[j]-‘0‘));

　　2、處理進位。乘積計算完後將進位由低位開始向高位進位調整每位符合十進制數

　　3、去掉前導零。

class Solution {
public:
    string multiply(string num1, string num2) {
        int len1=num1.length(),len2=num2.length();
        int len = len1+len2-2;
        int C = 0;
        int R[220]={0};
        
 
string ans;//對應位乘
        for(int i = 0 ; i < len1;i++){
            for(int j = 0;j < len2;j++){
                R[len - i - j]+=((num1[i]-‘0‘)*(num2[j]-‘0‘));
            }
        }
        //處理進位
        for(int i = 0 ; i < len1+len2 ;i++){
            R[i] += C;
            C = R[i]/10;
            R[i] %= 10;
        }
        //處理前導0
        int k = len1+len2;
        while(!R[k]){
            k--;
            if(k<0)
                return "0";
        }
        //壓入string返回
        for(int i = k ; i >= 0;i--){
            ans.push_back(R[i]+‘0‘);
        }
        return ans;
    }
};

leetcode multiply-strings