# 43. Multiply Strings

[ **43. Multiply Strings**](https://leetcode.com/problems/multiply-strings/description/)

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 integer** directly.

**My Solutions:**

因为相乘的结果数字可能很大，所以不能按照乘法的正常方法来做（数字相乘再相加）。

![](https://3140251282-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-L9mNqXfPqoVkVgLGF9r%2F-LBg2XchVuGu2TldsjWj%2F-LBgC2AgzRy1ZbsMil8E%2Fimage.png?alt=media\&token=fbb39c3a-62ff-4ee2-b340-39c7d314ef3f)

```java
class Solution {
    public String multiply(String num1, String num2) {
        if (num1 == null || num2 == null || num1.length() == 0 || num2.length() == 0) return "0";
        if (num1.equals("0") || num2.equals("0")) return "0";

        int[] res = new int[num1.length() + num2.length()];
        
        for (int i = num1.length() - 1; i >= 0; i--) {

            for (int j = num2.length() - 1; j >= 0; j--) {
                int product = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
                // p1是靠左的index，p2是靠右的
                int p1 = i + j, p2 = p1 + 1;
                
                int sum = res[p2] + product; //sum需要加上当下位置的数
                res[p1] += sum / 10; //p1是左边的位置
                res[p2] = sum % 10; //p2是当下位置
            } 
        }
        
        StringBuilder sb = new StringBuilder();
        for (int i : res) {
            if (!(i == 0 && sb.length() == 0)) { // e.g.[0, 0, 1, 0, 3] 跳过leading zero
                sb.append(i);
            }
        }
        return sb.toString();
    }
}
```

&#x20;Time: O(n \* m); Space: (n + m)
