[LeetCode]43. Multiply Strings

43. Multiply Strings

这道题要做的是将两个写成了string的数字相乘，思路很清晰，就是手算乘法的式子。我的想法是把这个过程模块化，写作一个字符乘以一个string和错位相加两个部分，可能实力还有欠缺写的不是很简洁。

class Solution {
public:
    string multiply(string num1, string num2) {
        if (num1=="0"||num2=="0") return "0";
        string res = "", pdt = "";
        int len1 = num1.size(), len2 = num2.size();
        for (int i = len1-1; i >= 0; i--){
            pdt = strProduct(num1[i],num2);
            for (int k = 0; k < len1-1-i; k++) pdt+="0";
            res = strAdd(res, pdt);
        }
        return res;
    }

    string strProduct(char num1, string num2){
        string res = "";
        int len = num2.size();
        int base = 0, carry = 0;
        for (int i = len-1; i >= 0; i--){
            base = (num1-'0')*(num2[i]-'0')+carry;
            res += to_string(base%10);
            carry = base/10;
        }
        if (carry!=0) res+=to_string(carry);
        reverse(res.begin(), res.end());
        return res;
    }

    string strAdd(string num1, string num2){
        // num1 is shorter than num2
        string res = "";
        int len1 = num1.size(), len2 = num2.size();
        int base = 0, carry = 0;
        int i = len1-1, j = len2-1;
        while (i>=0 && j>=0){
            base = (num1[i]-'0')+(num2[j]-'0')+carry;
            res+=to_string(base%10);
            carry = base/10;
            i--;j--;
        }
        while (j>=0){
            base = num2[j]-'0'+carry;
            res+=to_string(base%10);
            carry = base/10;
            j--;
        }
        if (carry!=0) res+=to_string(carry);
        reverse(res.begin(), res.end());
        return res;
    }
};