> For the complete documentation index, see [llms.txt](https://liuyang89116.gitbook.io/my-leetcode-book/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://liuyang89116.gitbook.io/my-leetcode-book/post_chapter_2_math/problem_43_multiply_strings.md).

# Problem 43: Multiply Strings

> <https://leetcode.com/problems/multiply-strings/>

## 思路

* 这道题不允许先转成 Integer 然后再计算。还有一点是直接乘的话数据会溢出。
* 直接乘会溢出，所以每次都要两个single digit相乘，最大81，不会溢出。
* 比如385 *97, 就是个位 = 5* 7，十位 = 8 *7 + 5* 9 ，百位 = 3 *7 + 8* 9 …
* 可以每一位用一个 Int 表示，存在一个int\[]里面。

  这个数组最大长度是num1.len + num2.len，比如99 \* 99，最大不会超过10000，所以4位就够了。
* num1 的第 i 位和 num2 的第 j 位乘，结果在 num\[i + j + 1] 位
* 最后结果前面的0要清掉。

```java
public class Solution {
    public String multiply(String num1, String num2) {
        int len1 = num1.length(), len2 = num2.length();
        int[] num3 = new int[len1 + len2];
        int i, j, product, carrier;

        for (i = len1 - 1; i >= 0; i--) {
            carrier = 0;
            for (j = len2 - 1; j >= 0; j--) {
                product = carrier + num3[i + j + 1] + Character.getNumericValue(num1.charAt(i)) * Character.getNumericValue(num2.charAt(j));
                num3[i + j + 1] = product % 10;
                carrier = product / 10;
            }
            num3[i + j + 1] = carrier;
        }

        StringBuilder sb = new StringBuilder();
        i = 0;
        while (i < num3.length - 1 && num3[i] == 0) {
            i++;
        }
        while (i < num3.length) {
            sb.append(num3[i]);
            i++;
        }

        return sb.toString();

    }
}
```

## 易错点

1. 和加法不一样，乘法要把之前的结果也加上 `num3[i + j + 1]`

   ```java
   product = carrier + num3[i + j + 1] + Character.getNumericValue(num1.charAt(i)) * Character.getNumericValue(num2.charAt(j));
   ```

   原因很简单就是，乘法一位一位地要不停地往前面乘。而加法每位数只“见一次面”
2. 把剩下的进位进到下一位去

   ```java
   num3[i + j + 1] = carrier;
   ```

   这时候的 num3\[i + j + 1] 和之前的 num3\[i + j + 1] 是不一样的，以为这时候，j = -1
3. 把没用的 0 都去掉

   ```java
   i = 0;
   while (i < num3.length - 1 && num3[i] == 0) {
        i++;
   }
   ```
4. char 转 int

   ```java
   Character.getNumericValue(num1.charAt(i))
   ```
5. 理清思路，注意两次 carrier 赋值和归 0 的位置

   ```java
   for (i = num1.length() - 1; i >= 0; i--) {
          carrier = 0;
          for (j = num2.length() - 1; j >= 0; j--) {
               product = carrier + rst[i + j + 1] + Character.getNumericValue(num1.charAt(i)) * Character.getNumericValue(num2.charAt(j));
               rst[i + j + 1] = product % 10;
               carrier = product / 10;
          }
          rst[i + j + 1] = carrier;
   }
   ```

   每换一次乘数，carrier 就要清 0.

## Follow Up

> 给一个数组，都是整型，比如\[1, 2, 3]。取每个数乘以2, 输出结果数组。有点类似string mulplication。比如 \[5, 5, 5] 输出 \[1, 1, 1, 0]. Follow up: 如果不reverse结果数组，要怎么做。回答用queue。

```java
/**
 * Created by yang on 1/3/17.
 */
public class Solution6 {
    public static int[] multiplyArray(int[] nums) {
        if (nums == null || nums.length == 0) {
            return null;
        }

        int[] rst = new int[nums.length];
        int digit = 0, carrier = 0;
        for (int i = nums.length - 1; i >= 0; i--) {
            int product = 2 * nums[i];
            digit = carrier + product % 10;
            carrier = product / 10;
            rst[i] = digit;
        }

        if (carrier == 0) return rst;
        int[] newArr = new int[nums.length + 1];
        for (int i = newArr.length - 1; i >= 0; i--) {
            if (i == 0) {
                newArr[0] = carrier;

            } else {
                newArr[i] = rst[i - 1];
            }
        }

        return newArr;
    }

    public static void main(String[] args) {
        int[] nums = new int[]{6, 9, 3};
        int[] rst = multiplyArray(nums);
        for (int i : rst) {
            System.out.print(i + " ");
        }
    }
}
```


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