leetcode [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.

 

题目大意：

求解两个string字符数字相乘得到的结果

 

解法：

首先，两个长度为m和n的数字相乘，结果的长度一定不会超过m+n，要么是m+n，要么是m+n-1，要么是0。当num1和num2中有一个为0，那么结果就是0。

第一个数字上第i个数字和第二个数字的第j个数字相乘得到的结果会作用在结果的第i+j+1个数字上，然后前面的i+j个数字取决于进位。

C++：

class Solution {
public:
    string multiply(string num1, string num2) {
        if(num1=="0"||num2=="0") return "0";
        int m=num1.size(),n=num2.size();
        string res(m+n,'0');
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                int t=int(num1[i]-'0')*int(num2[j]-'0');
                int count=t/10;
                res[i+j+1]=res[i+j+1]+t%10;
                if(res[i+j+1]>'9'){
                    res[i+j+1]=(res[i+j+1]-'0')%10+'0';
                    count++;
                } 
                int index=i+j;
                while(index>=0&&count>0){
                    res[index]=res[index]+count;
                    if(res[index]>'9'){
                        res[index]=(res[index]-'0')%10+'0';
                        count=1;
                    }else{
                        count=0;
                    }
                    index--;
                }
            }
        }

        return res[0]=='0'?res.substr(1):res;
    }
};