LeetCode 43. Multiply Strings

注意/思路

• 保存 被乘数*[0…9]的结果备用，然后移位相加即可
• 乘数末尾是0

代码

9ms

class Solution {
public:
    void ReverseStr(string& str){
        int left=0, right = str.size()-1;
        while(left<right){
            char c = str[left];
            str[left] = str[right];
            str[right] = c;
            left++;
            right--;
        }
    }
    void AddResult(string& result, const string& subProduct, const int shift){
        int carry = 0;
        while(result.size() < shift){
            result += "0";
        }
        for(int i=shift; i<subProduct.size() + shift; i++){
            if(i < result.size()){
                int subSum = (result[i]-'0') + (subProduct[i-shift]-'0') + carry;
                if(subSum >= 10){
                    subSum-= 10;
                    carry = 1;
                } else {
                    carry = 0;
                }
                result[i] = subSum + '0';
            } else {
                int subSum = (subProduct[i-shift]-'0') + carry;
                carry = subSum / 10;
                subSum = subSum % 10;
                result = result + (char)(subSum+'0');
            }
        }
        if (carry > 0) {
			result += (char)(carry + '0');
		}
    }
    string getSubProduct(char digit, string op){
        int carry = 0;
        int d = digit-'0';
        for(int i=0; i<op.size(); i++){
            int prod = (op[i]-'0') * d + carry;
            carry = prod / 10;
            prod = prod % 10;
            op[i] = '0' + prod;
        }
        if(carry > 0){
            op = op + (char)(carry+'0');
        }
        return op;
    }
    string multiply(string op1, string op2) {
        string result;
        if(op1.empty() || op2.empty())
            return result;
        if(op1.size() < op2.size()){
            string temp = op1;
            op1 = op2;
            op2 = temp;
        }
        string productSub[10];
        ReverseStr(op1);
        ReverseStr(op2);
        for(int i=0; i<op2.size(); i++){
            char digit = op2[i];
            if(digit == '0')
                continue;
            if(productSub[digit-'0'].empty()){
                productSub[digit-'0'] = getSubProduct(digit, op1);
            }
            AddResult(result, productSub[digit-'0'], i);
        }
        if(result.empty())
            return "0";
        ReverseStr(result);
        return result;
    }
};