leetcode -- 43. Multiply Strings

最新推荐文章于 2019-03-13 23:54:21 发布

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最新推荐文章于 2019-03-13 23:54:21 发布

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分类专栏：算法文章标签： leetcode

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本文详细解析了一种解决字符串形式的大数乘法问题的算法，该算法不使用内置的BigInteger库，也不直接将输入转换为整数，适用于长度小于110的非负整数字符串。通过逐位相乘并进位处理，最终返回两个数的乘积作为字符串输出。

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题目描述

题目难度：Medium
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.

AC代码

class Solution {
    public String multiply(String num1, String num2) {
    int m = num1.length(), n = num2.length();
    int[] pos = new int[m + n];
   
    for(int i = m - 1; i >= 0; i--) {
        for(int j = n - 1; j >= 0; j--) {
            int mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0'); 
            int p1 = i + j, p2 = i + j + 1;
            int sum = mul + pos[p2];

            pos[p1] += sum / 10;
            pos[p2] = (sum) % 10;
        }
    }  
    
    StringBuilder sb = new StringBuilder();
    for(int p : pos) if(!(sb.length() == 0 && p == 0)) sb.append(p);
    return sb.length() == 0 ? "0" : sb.toString();
}
}