LeetCode: 43. Multiply Strings

友情支持

如果您觉得这个笔记对您有所帮助，看在D瓜哥码这么多字的辛苦上，请友情支持一下，D瓜哥感激不尽，😜

支付宝

有些打赏的朋友希望可以加个好友，欢迎关注D 瓜哥的微信公众号，这样就可以通过公众号的回复直接给我发信息。

wx jikerizhi

公众号的微信号是: jikerizhi。因为众所周知的原因，有时图片加载不出来。如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。

43. Multiply Strings

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:

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

解题思路

将每次相乘的计算结果存入到一个数组中，每个元素保存一个位上的数字。

参考资料

优化版竖式(打败99.4%) - 字符串相乘 - 力扣（LeetCode） — 第二种解法，最初没看明白，但是受他们的启发，我自己想到了。

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:

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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
/**
 * Runtime: 5 ms, faster than 59.85% of Java online submissions for Multiply Strings.
 * Memory Usage: 38.5 MB, less than 16.67% of Java online submissions for Multiply Strings.
 */
public String multiply(String num1, String num2) {
    if ("0".equals(num1) || "0".equals(num2)) {
        return "0";
    }
    int[] products = new int[num1.length() + num2.length()];
    for (int i = num1.length() - 1; i >= 0; i--) {
        int iNum = num1.charAt(i) - '0';
        if (iNum == 0) {
            continue;
        }
        for (int j = num2.length() - 1; j >= 0; j--) {
            int jNum = num2.charAt(j) - '0';
            int sum = products[i + j + 1] + iNum * jNum;
            products[i + j + 1] = sum % 10;
            products[i + j] += sum / 10;
        }
    }
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < products.length; i++) {
        if (i == 0 && products[i] == 0) {
            continue;
        }
        sb.append(products[i]);
    }
    return sb.toString();
}