LeetCode - 43. Multiply Strings
43. Multiply Strings #
Problem's Link
#
----------------------------------------------------------------------------
Mean:
给定两个字符串,计算这两个字符串相乘的结果.
analyse:
模拟大数乘法.
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-06-17.13
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <windows.h>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
class Solution
{
public:
string multiply(string num1, string num2)
{
reverse(num1.begin(),num1.end());
reverse(num2.begin(),num2.end());
string res="0";
int begin=0;
for(int i=0;i<num2.length();++i)
{
string temp_res;
str_num_multi(num1,num2[i]-'0',temp_res);
add(res,temp_res,begin);
begin++;
}
reverse(res.begin(),res.end());
// delete leader zero
while(res.size()>0 && res[0]=='0')
{
res.pop_back();
}
if(res.size()==0) res.push_back(char('0'));
return res;
}
void str_num_multi(string str,int num,string& res)
{
int carry=0;
for(int i=0;i<str.length();++i)
{
int now=num*(str[i]-'0')+carry;
carry=now/10;
now%=10;
res.push_back(char(now+'0'));
}
if(carry)
res.push_back(char(carry+'0'));
}
void add(string& res,string num,int begin)
{
int len=num.length(),carry=0,now;
for(int i=0;i<len;++i)
{
if(begin<res.size())
now=(res[begin]-'0')+(num[i]-'0')+carry;
else now=carry+(num[i]-'0');
carry=now/10;
now%=10;
if(begin>=res.size())
res.push_back(char(now+'0'));
else
res[begin]=now+'0';
++begin;
}
while(carry)
{
if(begin<res.size())
now=carry+(res[begin]-'0');
else now=carry;
carry=now/10;
now%=10;
if(begin>=res.size())
{
res.push_back(char(now+'0'));
}
else
res[begin]=now+'0';
++begin;
}
}
};
int main()
{
string s1,s2;
while(cin>>s1>>s2)
{
Solution solution;
string ans=solution.multiply(s1,s2);
cout<<ans<<endl;
}
return 0;
}
/*
*/
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-06-17.13
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <windows.h>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
class Solution
{
public:
string multiply(string num1, string num2)
{
reverse(num1.begin(),num1.end());
reverse(num2.begin(),num2.end());
string res="0";
int begin=0;
for(int i=0;i<num2.length();++i)
{
string temp_res;
str_num_multi(num1,num2[i]-'0',temp_res);
add(res,temp_res,begin);
begin++;
}
reverse(res.begin(),res.end());
// delete leader zero
while(res.size()>0 && res[0]=='0')
{
res.pop_back();
}
if(res.size()==0) res.push_back(char('0'));
return res;
}
void str_num_multi(string str,int num,string& res)
{
int carry=0;
for(int i=0;i<str.length();++i)
{
int now=num*(str[i]-'0')+carry;
carry=now/10;
now%=10;
res.push_back(char(now+'0'));
}
if(carry)
res.push_back(char(carry+'0'));
}
void add(string& res,string num,int begin)
{
int len=num.length(),carry=0,now;
for(int i=0;i<len;++i)
{
if(begin<res.size())
now=(res[begin]-'0')+(num[i]-'0')+carry;
else now=carry+(num[i]-'0');
carry=now/10;
now%=10;
if(begin>=res.size())
res.push_back(char(now+'0'));
else
res[begin]=now+'0';
++begin;
}
while(carry)
{
if(begin<res.size())
now=carry+(res[begin]-'0');
else now=carry;
carry=now/10;
now%=10;
if(begin>=res.size())
{
res.push_back(char(now+'0'));
}
else
res[begin]=now+'0';
++begin;
}
}
};
int main()
{
string s1,s2;
while(cin>>s1>>s2)
{
Solution solution;
string ans=solution.multiply(s1,s2);
cout<<ans<<endl;
}
return 0;
}
/*
*/
作者:北岛知寒
出处:https://www.cnblogs.com/crazyacking/p/5248220.html
版权:本作品采用「署名-非商业性使用-相同方式共享 4.0 国际」许可协议进行许可。
分类:
ACM/LC
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 如何编写易于单元测试的代码
· 10年+ .NET Coder 心语,封装的思维:从隐藏、稳定开始理解其本质意义
· .NET Core 中如何实现缓存的预热?
· 从 HTTP 原因短语缺失研究 HTTP/2 和 HTTP/3 的设计差异
· AI与.NET技术实操系列:向量存储与相似性搜索在 .NET 中的实现
· 周边上新:园子的第一款马克杯温暖上架
· Open-Sora 2.0 重磅开源!
· 分享 3 个 .NET 开源的文件压缩处理库,助力快速实现文件压缩解压功能!
· Ollama——大语言模型本地部署的极速利器
· DeepSeek如何颠覆传统软件测试?测试工程师会被淘汰吗?