leetcode 166. Fraction to Recurring Decimal

/*
leetcode 166. Fraction to Recurring Decimal

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)".
Show Hint

题目大意：用小数表示除法的结果，如果是循环的小数，那么用括号表示循环的部分
解题思路：
1. 先得到整数部分
2. *10之后再除得到小数部分
3. 判断循环的方法：用一个hashmap记录每一个余数，当余数再次出现时就进入循环
注意：
1. 符号的判断
2. 当负数转为正数时可能会溢出
*/

#include <iostream>
#include <vector>
#include <string>
#include <unordered_map>

using namespace std;

class Solution {
public:
string fractionToDecimal(int numerator, int denominator)
{
if (numerator == 0)
return "0";

string ret;
//decide the sign of the result
if ((numerator > 0) ^ (denominator > 0))
ret += '-';  // the sign is '-'
// ret =  k * denominator + r;
long n = numerator;
long d = denominator;
n = abs(n);
d = abs(d);
long k = n / d;
long r = n % d;
ret += to_string(k);
if (!r)
return ret;
else
ret += ".";
//处理小数部分
unordered_map<long, int> umap;
while (r)
{
if (umap.find(r) != umap.end())
{
int pos = umap[r];
string s1 = ret.substr(0, pos);
string s2 = ret.substr(pos, ret.size());
ret = s1 + "(" + s2 + ")";
break;
}

umap[r] = ret.size(); // 记录余数的位置

r *= 10;
ret += to_string(r / d);
r = r % d;
}

return ret;
}

//参考答案 0-ms
//https://discuss.leetcode.com/topic/50057/0ms-17-lines-readable-c-solution
string fractionToDecimal1(int numerator, int denominator)
{
if (!numerator)
return "0";
string ret;
long long num = llabs(numerator), deno = llabs(denominator);
if (numerator < 0 ^ denominator < 0)
ret += '-';
ret += to_string(num / deno);
num = num % deno;
if (!num)
return ret;
unordered_map<int, int> umap;
ret += '.';
while (!umap.count(num))
{
umap[num] = ret.size();
while ((num *= 10) < deno)
{
ret += '0';
umap[num] = ret.size();
}
ret += '0' + num / deno;
if (!(num = num%deno))
return ret;
}
ret.insert(umap[num], 1, '(');
ret.insert(ret.size(), 1, ')');

return ret;
}
};

void test()
{
Solution sol;
cout << sol.fractionToDecimal1(1, 6) << endl;
cout << sol.fractionToDecimal1(2, 1) << endl;
cout << sol.fractionToDecimal1(1, 3) << endl;
cout << sol.fractionToDecimal1(2, 7) << endl;
}

int main()
{
test();

return 0;
}