---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0043.Multiply%20Strings/README.md
tags:
    - 数学
    - 字符串
    - 模拟
---

<!-- problem:start -->

# [43. 字符串相乘](https://leetcode.cn/problems/multiply-strings)

[English Version](/solution/0000-0099/0043.Multiply%20Strings/README_EN.md)

## 题目描述

<!-- description:start -->

<p>给定两个以字符串形式表示的非负整数&nbsp;<code>num1</code>&nbsp;和&nbsp;<code>num2</code>，返回&nbsp;<code>num1</code>&nbsp;和&nbsp;<code>num2</code>&nbsp;的乘积，它们的乘积也表示为字符串形式。</p>

<p><strong>注意：</strong>不能使用任何内置的 BigInteger 库或直接将输入转换为整数。</p>

<p>&nbsp;</p>

<p><strong>示例 1:</strong></p>

<pre>
<strong>输入:</strong> num1 = "2", num2 = "3"
<strong>输出:</strong> "6"</pre>

<p><strong>示例&nbsp;2:</strong></p>

<pre>
<strong>输入:</strong> num1 = "123", num2 = "456"
<strong>输出:</strong> "56088"</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>1 &lt;= num1.length, num2.length &lt;= 200</code></li>
	<li><code>num1</code>&nbsp;和 <code>num2</code>&nbsp;只能由数字组成。</li>
	<li><code>num1</code>&nbsp;和 <code>num2</code>&nbsp;都不包含任何前导零，除了数字0本身。</li>
</ul>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一：数学乘法模拟

假设 $num1$ 和 $num2$ 的长度分别为 $m$ 和 $n$，则它们的乘积的长度最多为 $m + n$。

证明如下：

-   如果 $num1$ 和 $num2$ 都取最小值，那么它们的乘积为 ${10}^{m - 1} \times {10}^{n - 1} = {10}^{m + n - 2}$，长度为 $m + n - 1$。
-   如果 $num1$ 和 $num2$ 都取最大值，那么它们的乘积为 $({10}^m - 1) \times ({10}^n - 1) = {10}^{m + n} - {10}^m - {10}^n + 1$，长度为 $m + n$。

因此，我们可以申请一个长度为 $m + n$ 的数组，用于存储乘积的每一位。

从低位到高位，依次计算乘积的每一位，最后将数组转换为字符串即可。

注意判断最高位是否为 $0$，如果是，则去掉。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m + n)$。其中 $m$ 和 $n$ 分别为 $num1$ 和 $num2$ 的长度。

<!-- tabs:start -->

#### Python3

```python
class Solution:
    def multiply(self, num1: str, num2: str) -> str:
        if num1 == "0" or num2 == "0":
            return "0"
        m, n = len(num1), len(num2)
        arr = [0] * (m + n)
        for i in range(m - 1, -1, -1):
            a = int(num1[i])
            for j in range(n - 1, -1, -1):
                b = int(num2[j])
                arr[i + j + 1] += a * b
        for i in range(m + n - 1, 0, -1):
            arr[i - 1] += arr[i] // 10
            arr[i] %= 10
        i = 0 if arr[0] else 1
        return "".join(str(x) for x in arr[i:])
```

#### Java

```java
class Solution {
    public String multiply(String num1, String num2) {
        if ("0".equals(num1) || "0".equals(num2)) {
            return "0";
        }
        int m = num1.length(), n = num2.length();
        int[] arr = new int[m + n];
        for (int i = m - 1; i >= 0; --i) {
            int a = num1.charAt(i) - '0';
            for (int j = n - 1; j >= 0; --j) {
                int b = num2.charAt(j) - '0';
                arr[i + j + 1] += a * b;
            }
        }
        for (int i = arr.length - 1; i > 0; --i) {
            arr[i - 1] += arr[i] / 10;
            arr[i] %= 10;
        }
        int i = arr[0] == 0 ? 1 : 0;
        StringBuilder ans = new StringBuilder();
        for (; i < arr.length; ++i) {
            ans.append(arr[i]);
        }
        return ans.toString();
    }
}
```

#### C++

```cpp
class Solution {
public:
    string multiply(string num1, string num2) {
        if (num1 == "0" || num2 == "0") {
            return "0";
        }
        int m = num1.size(), n = num2.size();
        vector<int> arr(m + n);
        for (int i = m - 1; i >= 0; --i) {
            int a = num1[i] - '0';
            for (int j = n - 1; j >= 0; --j) {
                int b = num2[j] - '0';
                arr[i + j + 1] += a * b;
            }
        }
        for (int i = arr.size() - 1; i; --i) {
            arr[i - 1] += arr[i] / 10;
            arr[i] %= 10;
        }
        int i = arr[0] ? 0 : 1;
        string ans;
        for (; i < arr.size(); ++i) {
            ans += '0' + arr[i];
        }
        return ans;
    }
};
```

#### Go

```go
func multiply(num1 string, num2 string) string {
	if num1 == "0" || num2 == "0" {
		return "0"
	}
	m, n := len(num1), len(num2)
	arr := make([]int, m+n)
	for i := m - 1; i >= 0; i-- {
		a := int(num1[i] - '0')
		for j := n - 1; j >= 0; j-- {
			b := int(num2[j] - '0')
			arr[i+j+1] += a * b
		}
	}
	for i := len(arr) - 1; i > 0; i-- {
		arr[i-1] += arr[i] / 10
		arr[i] %= 10
	}
	i := 0
	if arr[0] == 0 {
		i = 1
	}
	ans := []byte{}
	for ; i < len(arr); i++ {
		ans = append(ans, byte('0'+arr[i]))
	}
	return string(ans)
}
```

#### TypeScript

```ts
function multiply(num1: string, num2: string): string {
    if (num1 === '0' || num2 === '0') {
        return '0';
    }
    const m: number = num1.length;
    const n: number = num2.length;
    const arr: number[] = Array(m + n).fill(0);
    for (let i: number = m - 1; i >= 0; i--) {
        const a: number = +num1[i];
        for (let j: number = n - 1; j >= 0; j--) {
            const b: number = +num2[j];
            arr[i + j + 1] += a * b;
        }
    }
    for (let i: number = arr.length - 1; i > 0; i--) {
        arr[i - 1] += Math.floor(arr[i] / 10);
        arr[i] %= 10;
    }
    let i: number = 0;
    while (i < arr.length && arr[i] === 0) {
        i++;
    }
    return arr.slice(i).join('');
}
```

#### Rust

```rust
impl Solution {
    pub fn multiply(num1: String, num2: String) -> String {
        if num1 == "0" || num2 == "0" {
            return String::from("0");
        }
        let (num1, num2) = (num1.as_bytes(), num2.as_bytes());
        let (n, m) = (num1.len(), num2.len());
        let mut res = vec![];
        for i in 0..n {
            let a = num1[n - i - 1] - b'0';
            let mut sum = 0;
            let mut j = 0;
            while j < m || sum != 0 {
                if i + j == res.len() {
                    res.push(0);
                }
                let b = num2.get(m - j - 1).unwrap_or(&b'0') - b'0';
                sum += a * b + res[i + j];
                res[i + j] = sum % 10;
                sum /= 10;
                j += 1;
            }
        }
        res.into_iter()
            .rev()
            .map(|v| char::from(v + b'0'))
            .collect()
    }
}
```

#### C#

```cs
public class Solution {
    public string Multiply(string num1, string num2) {
        if (num1 == "0" || num2 == "0") {
            return "0";
        }

        int m = num1.Length;
        int n = num2.Length;
        int[] arr = new int[m + n];

        for (int i = m - 1; i >= 0; i--) {
            int a = num1[i] - '0';
            for (int j = n - 1; j >= 0; j--) {
                int b = num2[j] - '0';
                arr[i + j + 1] += a * b;
            }
        }

        for (int i = arr.Length - 1; i > 0; i--) {
            arr[i - 1] += arr[i] / 10;
            arr[i] %= 10;
        }

        int index = 0;
        while (index < arr.Length && arr[index] == 0) {
            index++;
        }

        StringBuilder ans = new StringBuilder();
        for (; index < arr.Length; index++) {
            ans.Append(arr[index]);
        }

        return ans.ToString();
    }
}
```

#### PHP

```php
class Solution {
    /**
     * @param string $num1
     * @param string $num2
     * @return string
     */

    function multiply($num1, $num2) {
        $length1 = strlen($num1);
        $length2 = strlen($num2);
        $product = array_fill(0, $length1 + $length2, 0);

        for ($i = $length1 - 1; $i >= 0; $i--) {
            for ($j = $length2 - 1; $j >= 0; $j--) {
                $digit1 = intval($num1[$i]);
                $digit2 = intval($num2[$j]);

                $temp = $digit1 * $digit2 + $product[$i + $j + 1];
                $product[$i + $j + 1] = $temp % 10;

                $carry = intval($temp / 10);
                $product[$i + $j] += $carry;
            }
        }
        $result = implode('', $product);
        $result = ltrim($result, '0');
        return $result === '' ? '0' : $result;
    }
}
```

#### Kotlin

```kotlin
class Solution {
    fun multiply(num1: String, num2: String): String {
        if (num1 == "0" || num2 == "0") return "0"

        val chars_1 = num1.toCharArray().reversedArray()
        val chars_2 = num2.toCharArray().reversedArray()

        val result = mutableListOf<Int>()

        chars_1.forEachIndexed { i, c1 ->
            val multiplier_1 = c1 - '0'
            var over = 0
            var index = 0

            fun sum(product: Int = 0): Unit {
                while (index >= result.size) {
                    result.add(0)
                }
                val value = product + over + result[index]
                result[index] = value % 10
                over = value / 10
                return
            }

            chars_2.forEachIndexed { j, c2 ->
                index = i + j
                val multiplier_2 = c2 - '0'
                sum(multiplier_1 * multiplier_2)
            }

            while (over > 0) {
                index++
                sum()
            }
        }

        return result.reversed().joinToString("")
    }
}
```

#### JavaScript

```js
/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
var multiply = function (num1, num2) {
    if (num1 === '0' || num2 === '0') return '0';

    const result = Array(num1.length + num2.length).fill(0);
    const code_0 = '0'.charCodeAt(0);

    const num1_len = num1.length;
    const num2_len = num2.length;

    for (let i = 0; i < num1_len; ++i) {
        const multiplier_1 = num1.charCodeAt(num1_len - i - 1) - code_0;
        for (let j = 0; j < num2_len; ++j) {
            const multiplier_2 = num2.charCodeAt(num2_len - j - 1) - code_0;
            result[i + j] += multiplier_1 * multiplier_2;
        }
    }

    result.reduce((carry, value, index) => {
        const sum = carry + value;
        result[index] = sum % 10;
        return (sum / 10) | 0;
    }, 0);

    return result
        .slice(0, result.findLastIndex(d => d !== 0) + 1)
        .reverse()
        .join('');
};
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->