Leetcode--Multiply Strings
2014-08-01 15:35
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Problem Description:
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:两个string相乘,主要是细节问题,具体方法是依次将一个字符串的每一位和另一个字符串的每一位相乘,要注意每次乘完一个字符后需要将起点加1,在结果中我是从第一位开始存的,完了之后将最终结果string转置即可。具体代码如下:
class Solution {
public:
string multiply(string num1, string num2) {
if(num1.empty()||num2.empty())
return "";
if(num1=="0"||num2=="0")
return "0";
int n1=num1.size();
int n2=num2.size();
string res(n1+n2,'0');//记录乘积结果
int flag=0,k=0;
for(int i=n1-1,start=0;i>=0;i--,start++)
{
k=0;
flag=0;
for(int j=n2-1;j>=0;j--)//num1的每一位依次和num2相乘,start表示每次相乘的起点
{
char temp='0'+((res[start+k]-'0')+(num2[j]-'0')*(num1[i]-'0')+flag)%10;
flag=((res[start+k]-'0')+(num2[j]-'0')*(num1[i]-'0')+flag)/10;
res[start+k]=temp;
k++;
}
if(flag!=0)
res[start+k]=('0'+flag);
}
if(flag!=0)
res[n1+n2-1]=('0'+flag);
res.assign(res.rbegin(),res.rend());//将最后的结果转置
if(res[0]=='0')
res=res.substr(1);
return res;
}
};
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:两个string相乘,主要是细节问题,具体方法是依次将一个字符串的每一位和另一个字符串的每一位相乘,要注意每次乘完一个字符后需要将起点加1,在结果中我是从第一位开始存的,完了之后将最终结果string转置即可。具体代码如下:
class Solution {
public:
string multiply(string num1, string num2) {
if(num1.empty()||num2.empty())
return "";
if(num1=="0"||num2=="0")
return "0";
int n1=num1.size();
int n2=num2.size();
string res(n1+n2,'0');//记录乘积结果
int flag=0,k=0;
for(int i=n1-1,start=0;i>=0;i--,start++)
{
k=0;
flag=0;
for(int j=n2-1;j>=0;j--)//num1的每一位依次和num2相乘,start表示每次相乘的起点
{
char temp='0'+((res[start+k]-'0')+(num2[j]-'0')*(num1[i]-'0')+flag)%10;
flag=((res[start+k]-'0')+(num2[j]-'0')*(num1[i]-'0')+flag)/10;
res[start+k]=temp;
k++;
}
if(flag!=0)
res[start+k]=('0'+flag);
}
if(flag!=0)
res[n1+n2-1]=('0'+flag);
res.assign(res.rbegin(),res.rend());//将最后的结果转置
if(res[0]=='0')
res=res.substr(1);
return res;
}
};
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