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

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contains only digits 0-9.
3. Both num1 and num2 does not contain any leading zero.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

　　基本思路：如上圖所示，因為兩個字符串相乘，長度不定，可以將結果存到一個數組中。我們可以看到num1[i]和num2[j]的乘積最後位置是在indices[i+j]和indices[i+j+1]兩個位置上，同時我們知道

一個m位的數乘一個n位的數做多是m+n位。

 1 class Solution {
 2 public:
 3     string multiply(string num1, string num2) {//num1[i] * num2[j]` will be placed         
 4                                                //
 
at indices `[i + j`, `i + j + 1]` m*n位數最多m+n位
 5         string result;
 6         int k = 0, sum = 0;
 7         int len1 = num1.size(), len2 = num2.size();
 8         vector<int>  res(len1 + len2);
 9         for (int i = len1-1; i>=0;i--)
10         {
11             for (int j = len2-1; j >= 0
 
; j--)
12             {
13                 int k = (num1[i] - ‘0‘)*(num2[j]-‘0‘);
14                 sum = k + res[i + j + 1];//前一次最高位
15                 res[i + j + 1] = sum % 10;
16                 res[i + j] += sum / 10;
17             }
18         }
19         //去除首位的0
20         int j = len1 + len2 - 1;
21         for (int i = 0; i <len1+len2; i++)
22         {
23             if (res[i] == 0) continue;
24             else
25             {
26                 j = i;
27                 break;
28             }
29         }
30         for (int i = j; i < len1 + len2; i++)
31         {
32             result += res[i] + ‘0‘;
33         }
34         return result;
35     }
36 };

Multiply Strings