LeetCode 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)".

 

Solution:

 1    string fractionToDecimal(int numerator, int denominator) {
 2         int neg = 0;
 3         int flag = 0;
 4         int recurring = 1;
 5          if ((numerator < 0 && denominator > 0) || (numerator > 0 && denominator < 0))  neg = 1;
 6         unsigned long long num = (numerator == -2147483648)? 2147483648 : abs(numerator);
 7         unsigned long long den = (denominator == -2147483648)? 2147483648 : abs(denominator);
 8     
 9         unsigned long long integer = num / den;
10         unsigned long long remain = num % den;
11         flag = (remain == 0 ? 1 : 0);
12     
13         map<int, int> seen;
14         seen[remain] = 0;
15         vector<int> digits;
16         while (1) {
17             remain *= 10;
18             digits.push_back(remain / den);
19             remain = remain % den;
20             recurring = (remain == 0) ? 0 : 1;
21             if (seen.count(remain)) {
22                 int where = seen[remain];
23                 ostringstream os;
24                 if (neg) {
25                     os << '-';
26                 }
27                 os << integer;
28                 if (!flag) {
29                     os << '.';
30                     for (int i = 0; i < where; ++i)
31                     {
32                         os << digits[i];
33                     }
34                     if (recurring)
35                     {
36                          os << '(';
37                         for (int i = where; i < digits.size(); ++i) 
38                             os << digits[i];
39                         os <<')';
40                     }
41                    
42                 }
43     
44                 return os.str();
45             }
46             else 
47                 seen[remain] = digits.size();
48     
49         }
50 
51     }

 

posted @ 2014-12-17 21:22  HaruHaru  阅读(216)  评论(0)  编辑  收藏  举报