LeetCode #43 - Multiply Strings

最新推荐文章于 2021-03-29 02:38:32 发布

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本文介绍了一种不使用内置大数库或直接转换输入为整数的方法，实现两个大数字符串的乘法运算。通过逐位相乘并处理进位，最终返回乘积的字符串形式。适用于长度不超过110的数字字符串。

题目描述：

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:

1. The length of both num1 and num2 is < 110.

2. Both num1 and num2 contain only digits 0-9.

3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.

4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

一个数字的第i位乘一个数字的第j位，就会加在积的第(i+j)位上，所以先计算积的每一位的累计结果，然后再从最低位开始，发现大于9，就向高位进位。

class Solution {
public:
    string multiply(string num1, string num2) {
        int n1=num1.size();
        int n2=num2.size();
        vector<int> v(n1+n2,0);
        int k=n1+n2-2;  // i+j最多为n1+n2-2
        for(int i=0;i<n1;i++)
            for(int j=0;j<n2;j++)
                v[k-i-j]+=(num1[i]-'0')*(num2[j]-'0');
        
        int carry=0;
        for(int i=0;i<n1+n2;i++)
        {
            v[i]+=carry;
            carry=v[i]/10;
            v[i]%=10;
        }
        string result;
        int i=n1+n2-1;
        while(v[i]==0) i--;
        if(i<0) return "0";
        while(i>=0)
        {
            result=result+to_string(v[i]);
            i--;
        }
        return result;
    }
};