【LeetCode练习题】Multiply Strings
Multiply Strings
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.
题目意思:
给两个string,计算string的乘积。
string中的数可以非常大并且是非负数。
(就是大数的乘法运算嘛。。。)
解题思路:
由于之前已经练习过了大数的加减乘除四则运算了,所以这道题跟那个是一样的原理。
要想实现乘法,首先得实现大数的加法,因为按照小学摆竖式计算乘法的时候需要在会加法的基础上才行。
于是我们首先实现了大数的加法,基本上就是模拟了用竖式每一位相加的过程,其中注意到产生的进制要参与高位的运算。
实现了加法后,乘法也就出来了。
代码如下:
string operator+(const string &num1,const string &num2){
int nCarry = ;
string numTemp;
int i = num2.size() - ;
int j = num1.size() - ;
for(; i >= || j >= ; i--,j--){
char a,b;
if(i>=)
b = num2[i] - '';
else
b = ;
if(j>=)
a = num1[j] - '';
else
a = ;
char c = a + b + nCarry;
nCarry = c / ;
numTemp.push_back(c% + '');
}
if(nCarry != ){
numTemp.push_back(nCarry + '');
}
for(i = , j = numTemp.size() - ; i < j; i++,j--){
char cTemp = numTemp[i];
numTemp[i] = numTemp[j];
numTemp[j] = cTemp;
}
return numTemp;
}
string operator*(const string &num1,const string &num2){
int nCarry = ;
string numTemp;
string result;
int i = num2.size()-;
for(; i >= ; i--){
char a = num2[i] - '';
int j = num1.size() - ;
for(; j >= ; j--){
char b = num1[j] - '';
char c = b * a + nCarry;
nCarry = c/;
numTemp.push_back(c % + '');
}
if(nCarry != ){
numTemp.push_back(nCarry + '');
nCarry = ;
}
//reverse
int n = ;
int m = numTemp.size() - ;
for(; n < m; n++,m--){
char cTemp = numTemp[n];
numTemp[n] = numTemp[m];
numTemp[m] = cTemp;
}
for(int t = num2.size() - ; t > i; t--)
{
numTemp.push_back('');
}
result = result + numTemp;
numTemp.clear();
}
return result;
}
class Solution {
public:
string multiply(string num1, string num2) {
string ret;
string zero = "";
if(!num1.compare(zero) || !num2.compare(zero)){
ret = "";
return ret;
}
ret = num1 * num2;
return ret;
}
};
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