43. 字符串相乘

题目

43. 字符串相乘

我的思路

我的思路是用num1的每一位去乘num2，产生num1长度个字符串，再把k个字符串相加。

我的实现

class Solution {
public:
    string multiply(string num1, string num2) {
        int l1 = num1.size();
        int l2 = num2.size();
        vector<string> products;
        for(int i=l1-1;i>=0;i--){
            string temp = "";
            for(int num0 = 1;num0<l1-i;num0++){
                temp += '0'; 
            }
            int o = 0;
            int pro = 0;
            int n = 0;
            for(int j=l2-1;j>=0;j--){
                pro = (num1[i] - '0')*(num2[j] - '0');
                n = (pro+o)%10;
                o = (pro+o)/10;
                temp += (n+'0');
                //cout<<"check"<<l2<<endl;
            }
            if(o>0) temp+= (o+'0');
            //reverse(temp.begin(),temp.end());
            products.push_back(temp);
            //cout<<temp<<endl;
        }
        //把得到的字符串相加
        string result = "";
        int preSum = 0;
        int nowSum = 0;
        for(int i=0;i<l1+l2;i++){
            int k=0;
            //preSum = 0;
            nowSum = 0;
            for(int k =0;k<l1;k++){
                if(i<products[k].size()){
                    nowSum += products[k][i] - '0';
                }
            }
            nowSum = nowSum + preSum;
            //if(nowSum>0||i==0)
            result += '0' + nowSum%10;
            preSum = nowSum/10;
            //cout<<preSum<<endl;
        }
        while(preSum>0){
            result += '0' + preSum%10;
            preSum = preSum/10;
        }
        int i = result.size()-1;
        while(*(result.end()-1)=='0'&&(result.end()-1)!=result.begin())result.erase(result.end()-1);
        //for()
        reverse(result.begin(),result.end());
        return result;
    }
};
/*
数字过长，无法用普通数据类型表示。
用num1的每一位去乘num2，产生x1个字符串，再把k个字符串相加。
*/

拓展学习

官方题解的思路更优雅

 

class Solution {
public:
    string multiply(string num1, string num2) {
        if (num1 == "0" || num2 == "0") {
            return "0";
        }
        int m = num1.size(), n = num2.size();
        auto ansArr = vector<int>(m + n);
        for (int i = m - 1; i >= 0; i--) {
            int x = num1.at(i) - '0';
            for (int j = n - 1; j >= 0; j--) {
                int y = num2.at(j) - '0';
                ansArr[i + j + 1] += x * y;
            }
        }
        for (int i = m + n - 1; i > 0; i--) {
            ansArr[i - 1] += ansArr[i] / 10;
            ansArr[i] %= 10;
        }
        int index = ansArr[0] == 0 ? 1 : 0;
        string ans;
        while (index < m + n) {
            ans.push_back(ansArr[index]);
            index++;
        }
        for (auto &c: ans) {
            c += '0';
        }
        return ans;
    }
};

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/multiply-strings/solution/zi-fu-chuan-xiang-cheng-by-leetcode-solution/
来源：力扣（LeetCode）
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