LeetCode - Fraction to Recurring Decimal

Fraction to Recurring Decimal

2015.1.23 15:59

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:

　　First of all, the denominator won't be 0. Next you should take care of the sign first, + or -, make it positive.

　　3 % 5 == 3

　　5 % 3 == 2

　　6 % 3 == 0

　　The process of calculating a rational fraction is actually by division and modulus. If the remainder is zero, you get a finite fraction, otherwise an infinite loop. There's gonna be a loop as long as it is rational fraction. So, you'll have to record the remainder sequence, with a hash table maybe. When the same remainder is encountered, you know you've found the loop sequence.

　　Also there's something to be careful about printing. Like the following.

　　5 / 3 == 1.(6)

　　0 / 3 == 0

　　6 / 3 == 2

　　1 / 7 == 0.(142857)

　　The denominator can be really large, so don't try to use an int A[MAXN] to record the sequence, it won't do. That's why I suggest hashing.

　　The time complexity is generally O(denominator), but should be much smaller in fact. The fraction 1 / n may have a loop sequence of at most n - 1 in length, but usually far shorter. Space complexity is the same scale.

Accepted code:

 1 // 1RE, 1TLE, 3MLE, 1AC, watch out for boundary cases and memory limit

 2 #include <iostream>

 3 #include <string>

 4 #include <vector>

 5 #include <unordered_map>

 6 using namespace std;

 7 

 8 typedef long long int ll;

 9 

10 class Solution {

11 public:

12     string fractionToDecimal(int numerator, int denominator) {

13         ll n = numerator;

14         ll d = denominator;

15 

16         int sn, sd;

17 

18         sn = n >= 0 ? 1 : -1;

19         n = n >= 0 ? n : -n;

20         sd = d >= 0 ? 1 : -1;

21         d = d >= 0 ? d : -d;

22 

23         if (n == 0) {

24             return "0";

25         }

26 

27         string res = "";

28         if (sn * sd < 0) {

29             res.push_back('-');

30         }

31 

32         ll q = n / d;

33         ll r = n % d;

34 

35         res += to_string(q);

36 

37         if (r == 0) {

38             return res;

39         }

40 

41         res.push_back('.');

42         int pos = (int)res.length();

43 

44         unordered_map<ll, int> seq;

45         unordered_map<ll, int>::iterator it;

46 

47         while (r != 0) {

48             it = seq.find(r);

49             if (it != seq.end()) {

50                 res.insert(it->second, "(");

51                 res.push_back(')');

52                 return res;

53             }

54             seq[r] = pos;

55             q = r * 10 / d;

56             r = r * 10 % d;

57             res.push_back((int)q + '0');

58             ++pos;

59         }

60 

61         seq.clear();

62         return res;

63     }

64 private:

65 };

66 /*

67 int main()

68 {

69     int n, d;

70     Solution sol;

71 

72     while (cin >> n >> d) {

73         cout << sol.fractionToDecimal(n, d) << endl;

74     }

75 

76     return 0;

77 }

78 */