Given two integers representing the numerator and denominator of a fraction, return the fraction in string format.
If the fractional part is repeating, enclose the repeating part in parentheses.
For example,
- Given numerator = 1, denominator = 2, return "0.5".
- Given numerator = 2, denominator = 1, return "2".
- Given numerator = 2, denominator = 3, return "0.(6)".
解题思路：
首先得到商的整数部分，余数部分不能整除则在余数后面加0。用一个vector保存每次相除后的余数，出现了相同的余数，则意味着有循环小数。
一个容易忽略的点是INT_MIN，即0x80000000，它的整数用int保存不了，需要转换格式为long long。
实现代码：
/*****************************************************************************
* @COPYRIGHT NOTICE
* @Copyright (c) 2015, 楚兴
* @All rights reserved
* @Version : 1.0
* @Author : 楚兴
* @Date : 2015/2/6 19:57
* @Status : Accepted
* @Runtime : 18 ms
*****************************************************************************/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
if (denominator == 0)
{
return ""; //分母为0
}
if (numerator == 0)
{
return "0"; //分子为0
}
long long num = numerator;
long long den = denominator;
bool flag = false;
if ((num > 0 && den < 0) ||(num < 0 && den > 0))
{
flag = true;
}
num = num >= 0 ? num : -num;
den = den > 0 ? den : - den;
string str = "";
long long result = num / den; //商的整数部分
char res[11];
itoa(result,res);
int len = strlen(res);
str += res;
long long n = num % den;
if (!n)
{
if (flag)
{
str.insert(str.begin(),'-');
}
return str;
}
str += '.';
vector<long long> tail;
int index = -1;
while(n)
{
tail.push_back(n);
n *= 10;
str += n / den + '0';
n %= den;
index = find(tail,n);
if (index != -1)
{
break;
}
}
if (index != -1)
{
str.insert(str.begin() + index + len + 1, '(');
str.push_back(')');
}
if (flag)
{
str.insert(str.begin(),'-');
}
return str;
}
int find(vector<long long> num, long long n)
{
for (int i = 0; i < num.size(); i++)
{
if (num[i] == n)
{
return i;
}
}
return -1;
}
void itoa(long long n, char* ch) //n不为负
{
char* temp = ch;
if (n == 0)
{
*temp++ = '0';
*temp = '\0';
return;
}
while(n)
{
*temp++ = n % 10 + '0';
n /= 10;
}
*temp = '\0';
reverse(ch);
}
void reverse(char* ch)
{
char* start = ch;
char* end = ch + strlen(ch) - 1;
char c;
while(start < end)
{
c = *start;
*start++ = *end;
*end-- = c;
}
}
};
int main()
{
Solution s;
cout<<s.fractionToDecimal(-2147483648, 1).c_str()<<endl;
cout<<s.fractionToDecimal(-1,-2147483648).c_str()<<endl;
cout<<s.fractionToDecimal(1,6).c_str()<<endl;
cout<<s.fractionToDecimal(-6,-7).c_str()<<endl;
system("pause");
}
