[LeetCode] 167. Fraction to Recurring Decimal 分数转循环小数
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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.
Example 1:
Input: numerator = 1, denominator = 2 Output: "0.5"
Example 2:
Input: numerator = 2, denominator = 1 Output: "2"
Example 3:
Input: numerator = 2, denominator = 3 Output: "0.(6)"
给2个整数分别作分子和分母,返回分数的字符串形式。
Java:
public class Solution { public String fractionToDecimal(int numerator, int denominator) { if (numerator == 0) { return "0"; } StringBuilder res = new StringBuilder(); // "+" or "-" res.append(((numerator > 0) ^ (denominator > 0)) ? "-" : ""); long num = Math.abs((long)numerator); long den = Math.abs((long)denominator); // integral part res.append(num / den); num %= den; if (num == 0) { return res.toString(); } // fractional part res.append("."); HashMap<Long, Integer> map = new HashMap<Long, Integer>(); map.put(num, res.length()); while (num != 0) { num *= 10; res.append(num / den); num %= den; if (map.containsKey(num)) { int index = map.get(num); res.insert(index, "("); res.append(")"); break; } else { map.put(num, res.length()); } } return res.toString(); } }
Python:
class Solution(object): def fractionToDecimal(self, numerator, denominator): """ :type numerator: int :type denominator: int :rtype: str """ result = "" if (numerator > 0 and denominator < 0) or (numerator < 0 and denominator > 0): result = "-" dvd, dvs = abs(numerator), abs(denominator) result += str(dvd / dvs) dvd %= dvs if dvd > 0: result += "." lookup = {} while dvd and dvd not in lookup: lookup[dvd] = len(result) dvd *= 10 result += str(dvd / dvs) dvd %= dvs if dvd in lookup: result = result[:lookup[dvd]] + "(" + result[lookup[dvd]:] + ")" return result
C++:
// upgraded parameter types string fractionToDecimal(int64_t n, int64_t d) { // zero numerator if (n == 0) return "0"; string res; // determine the sign if (n < 0 ^ d < 0) res += ‘-‘; // remove sign of operands n = abs(n), d = abs(d); // append integral part res += to_string(n / d); // in case no fractional part if (n % d == 0) return res; res += ‘.‘; unordered_map<int, int> map; // simulate the division process for (int64_t r = n % d; r; r %= d) { // meet a known remainder // so we reach the end of the repeating part if (map.count(r) > 0) { res.insert(map[r], 1, ‘(‘); res += ‘)‘; break; } // the remainder is first seen // remember the current position for it map[r] = res.size(); r *= 10; // append the quotient digit res += to_string(r / d); } return res; }
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