# 分数到小数
给定两个整数,分别表示分数的分子 numerator
和分母 denominator
,以 字符串形式返回小数 。
如果小数部分为循环小数,则将循环的部分括在括号内。
如果存在多个答案,只需返回 任意一个 。
对于所有给定的输入,保证 答案字符串的长度小于 104
。
示例 1:
输入:numerator = 1, denominator = 2
输出:"0.5"
示例 2:
输入:numerator = 2, denominator = 1
输出:"2"
示例 3:
输入:numerator = 2, denominator = 3
输出:"0.(6)"
示例 4:
输入:numerator = 4, denominator = 333
输出:"0.(012)"
示例 5:
输入:numerator = 1, denominator = 5
输出:"0.2"
提示:
-231 <= numerator, denominator <= 231 - 1
denominator != 0
以下错误的选项是?
## aop
### before
```c
#include
using namespace std;
```
### after
```c
int main()
{
Solution sol;
string res;
int numerator = 1, denominator = 2;
res = sol.fractionToDecimal(numerator, denominator);
cout << res;
return 0;
}
```
## 答案
```c
class Solution
{
public:
const static int CACHE_SIZE = 1000;
string fractionToDecimal(int numerator, int denominator)
{
if (numerator == 0)
return "0";
bool isNegative = numerator < 0 ^ denominator < 0;
double doubleNumerator = fabs(numerator);
double doubleDenominator = fabs(denominator);
double integral = floor(doubleNumerator / doubleDenominator);
doubleNumerator -= doubleDenominator * integral;
int len = 0;
string cache;
vector dividendCache;
bool isRecurring = false;
int recurringStart = 0;
int i = 0;
while (doubleNumerator)
{
doubleNumerator *= 10;
len = dividendCache.size();
for (i = 0; i < len; ++i)
if (dividendCache[i] == doubleNumerator)
{
isRecurring = true;
recurringStart = i;
break;
}
if (isRecurring)
break;
i = (int)(floor(doubleNumerator / doubleDenominator));
cache += i;
dividendCache.push_back(doubleNumerator);
doubleNumerator -= doubleDenominator * i;
}
string ret = isNegative ? "-" : "";
string tmp;
char c;
if (integral == 0)
ret += "0";
else
{
while (integral)
{
c = (int)(integral - 10 * floor(integral / 10)) + '0';
tmp = c + tmp;
integral = floor(integral / 10);
}
ret += tmp;
}
if (dividendCache.size() > 0)
ret += '.';
i = 0;
if (isRecurring)
{
ret += cache.substr(0, recurringStart);
ret += '(';
ret += cache.substr(recurringStart);
ret += ')';
}
else
ret += cache;
return ret;
}
};
```
## 选项
### A
```c
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
typedef long long LL;
LL x = numerator, y = denominator;
if (x % y == 0)
return to_string(x / y);
string res;
if ((x < 0) ^ (y < 0))
res += '-';
x = abs(x), y = abs(y);
res += to_string(x / y) + '.';
x %= y;
unordered_map hash;
while (x)
{
hash[x] = res.size();
x *= 10;
res += to_string(x / y);
x %= y;
if (hash.count(x))
{
res = res.substr(0, hash[x]) + '(' + res.substr(hash[x]) + ')';
break;
}
}
return res;
}
};
```
### B
```c
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
string ans;
unordered_map use;
if (numerator > 0 && denominator < 0 || numerator < 0 && denominator > 0)
ans = '-';
long numerat = fabs(numerator);
long denominat = fabs(denominator);
ans += to_string(numerat / denominat);
numerat = numerat % denominat;
int point, x;
if (numerat)
{
ans += ".";
point = ans.length() - 1;
}
else
return ans;
while (numerat && use.count(numerat) == 0)
{
use[numerat] = ++point;
numerat *= 10;
x = numerat / denominat;
numerat = numerat % denominat;
ans.push_back(x + '0');
}
if (numerat)
{
ans.insert(ans.begin() + use[numerat], '(');
ans.push_back(')');
}
return ans;
}
};
```
### C
```c
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
int m1 = numerator > 0 ? 1 : -1;
int m2 = denominator > 0 ? 1 : -1;
long long num = abs((long long)numerator);
long long den = abs((long long)denominator);
long long t = num / den;
long long res = num % den;
string ans = to_string(t);
if (m1 * m2 == -1 && (t > 0 || res > 0))
ans = '-' + ans;
if (res == 0)
return ans;
ans += '.';
unordered_map m;
string s = "";
int pos = 0;
while (res != 0)
{
if (m.find(res) != m.end())
{
s.insert(m[res], "(");
s += ')';
return ans + s;
}
m[res] = pos;
s += to_string((res * 10) / den);
res = (res * 10) % den;
pos++;
}
return ans + s;
}
};
```