Multiply Strings

最新推荐文章于 2024-08-01 17:15:16 发布

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最新推荐文章于 2024-08-01 17:15:16 发布

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分类专栏： leetcode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/Ada_baby/article/details/51057556

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本文介绍了一种处理任意精度大数乘法的方法，通过字符串形式表示大数，并提供了一个C++实现示例。该方法避免了使用标准整数类型处理大数时可能出现的溢出问题。

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Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

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class Solution {
public:
    string multiply(string num1, string num2) {
        if(num1 == "0" || num2 == "0")
            return "0";
        int len1 = num1.size();
        int len2 = num2.size();

        int len = len1 + len2;
        int *iNum1 = new int[len1]();
        int *iNum2 = new int[len2]();
        int *result = new int[len1 + len2]();
        for(int i = 0; i < len1; i++) 
            iNum1[i] = num1[i] - '0';
        for(int i = 0; i < len2; i++)
            iNum2[i] = num2[i] - '0';
        for(int i = 0; i < len1; i++) {
            for(int j = 0; j < len2; j++) {
                result[i + j + 1] += iNum1[i] * iNum2[j];
            }
        }
        #if 0
        for(int i = 0; i < len ; i++)
            cout << result[i] << " ";
        cout << endl;
        #endif
        for(int i = len - 1; i> 0; i--) {
            result[i - 1] += result[i] / 10;
            result[i] %= 10;
        }
        #if 0
        for(int i = 0; i < len ; i++)
            cout << result[i] << " ";
        cout << endl;
        #endif
        string ret = "";
        for(int i = 0; i < len; i++)
            ret.append(1, result[i] + '0');
        if(ret[0] == '0')
            return ret.substr(1);
        return ret;
    }
};