【题目】

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

【分析】

高精度乘法（大数乘法）

【代码】

/*********************************
*   日期：2015-01-28
*   作者：SJF0115
*   题目: 43.Multiply Strings
*   网址：https://oj.leetcode.com/problems/multiply-strings/
*   结果：AC
*   来源：LeetCode
*   博客：
**********************************/
#include <iostream>
#include <cstring>
using namespace std;

class Solution {
public:
    string multiply(string num1, string num2) {
        int len1 = num1.length();
        int len2 = num2.length();
        // 容错处理
        if(len1 <= 0 || len2 <= 0){
            return "";
        }//if
        int sum = 0;
        int len3 = len1 + len2;
        char result[len3];
        memset(result,'0',sizeof(result[0])*(len3+1));
        for(int i = len1 - 1,m = 0;i >= 0;--i,++m){
            for(int j = len2 - 1,n = 0;j >= 0;--j,++n){
                sum = (num1[i] - '0') * (num2[j] - '0') + result[m+n] - '0';
                result[m+n] = sum % 10 + '0';
                // 进位
                result[m+n+1] += sum / 10;
            }//for
        }//for
        //确定乘积的位数
        while(result[len3] == '0' && len3 > 0){
            --len3;
        }//while
        //注意：加'\0'
        result[len3+1] = '\0';
        //翻转
        int temp;
        for(int i = 0,j = len3;i < j;++i,--j){
            temp = result[i];
            result[i] = result[j];
            result[j] = temp;
        }//for
        return string(result);
    }
};

int main(){
    Solution solution;
    string num1("0");
    string num2("123");
    string result = solution.multiply(num1,num2);
    // 输出
    cout<<result<<endl;
    return 0;
}

/*for(int i = 0;i < result.size();++i){
        for(int j = 0;j < result[i].size();++j){
            cout<<result[i][j]<<" ";
        }
        cout<<endl;
    }*/

[LeetCode]43.Multiply Strings