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 */

 

 posted on 2015-01-23 16:40  zhuli19901106  阅读(268)  评论(0编辑  收藏  举报