[LeetCode] 167. Fraction to Recurring Decimal 分数转循环小数
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;
}
