[LeetCode] 43. Multiply Strings

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本文介绍了一种不使用内置大整数库实现两个非负整数相乘的方法，通过字符串形式输入并返回乘积的字符串表示。文章对比了两种实现方式：一种是逐位相乘再累加；另一种是更高效的利用数组进行位运算。

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.

解析

实现字符串的乘法。自己写的时候是每一个字符去乘另一个字符得到字符串再相加，看了大神的解法后，觉得自己的太naive了。
在这里插入图片描述
两个数相乘得到的数的长度为两个数长度之和，并且两个数的位置相乘得到的数在i+j和i+j+1两个位置上。从后往前乘。

自己写的垃圾

class Solution {
public:
    string add(string num1, string num2){
        int len1 = num1.size();
        int len2 = num2.size();
        int i=0,j=0;
        int carry = 0;
        string res = "";
        while(i<len1 && j<len2){
            res +=  (((num1[i] -'0')+(num2[j]-'0')+carry)%10) + '0';
            carry = ((num1[i] -'0')+(num2[j]-'0')+ carry) / 10;
            i++;j++;
        }
        while(i<len1){
            res +=  (((num1[i] -'0')+carry)%10) + '0';
            carry = ((num1[i] -'0')+carry)/10;
            i++;
        }
        while(j<len2){
            res +=  (((num2[j] -'0')+carry)%10) + '0';
            carry = ((num2[j] -'0')+carry)/10;
            j++;
        }
        if(carry>0)
            res += (carry+'0');
        return res;
    }


    string multiply(string num1, string num2) {
        if(num1 == "0" || num2 == "0")
            return "0";
        int len1 = num1.size();
        int len2 = num2.size();
        if(len2 > len1){
            string tmp = num2;
            num2 = num1;
            num1 = tmp;
        }
        len1 = num1.size(); len2 = num2.size();
        string res = "";
        string carry = "";
        for(int i=len2-1;i>=0;i--){
            int carry1 = 0;
            string tmp = "";
            for(int j=len1-1;j>=0;j--){
                char ch = (((num2[i] -'0')*(num1[j]-'0') + carry1)%10) + '0';
                cout<<ch<<endl;
                tmp += (((num2[i] -'0')*(num1[j]-'0') + carry1)%10) + '0';
                carry1 = ((num2[i] -'0')*(num1[j]-'0')+ carry1)/10;
            }
            if(carry1>0)
                tmp += (carry1+'0');
            tmp = carry + tmp;
            carry += '0';
            res = add(res, tmp);
        }
        reverse(res.begin(),res.end());
        return res;
    }
};

大神解法

class Solution {
public:
    string multiply(string num1, string num2) {
        if(num1 == "0" || num2 == "0")
            return "0";
        int len1 = num1.size();
        int len2 = num2.size();
        string result = "";
        vector<int> res(len1+len2, 0);
        for(int i=len2-1;i>=0;i--){
            for(int j=len1-1;j>=0;j--){
                int mul = (num2[i]-'0')*(num1[j]-'0');
                res[i+j] += (mul+res[i+j+1])/10;
                res[i+j+1] = (mul+res[i+j+1])%10;
            }
        }
        for(int i=0;i<len1+len2;i++){
            if (!result.empty() || res[i] != 0) 
                result += (res[i] + '0');
        }
        return result;
    }
};