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)".
把整数相除的结果用string表示,小数的循环部分用()框住。先把整数部分输出到string,然后模拟人类算小数的步骤,每次除数乘以10除以被除数得到的余数成为新的除数,当遇到见过的被除数时意味着循环出现,加()并退出。当被除数是2^(-31)时除数会一直乘以10到溢出,因此除数和被除数换成更大的数据类型。
/*
* @lc app=leetcode id=166 lang=cpp
*
* [166] Fraction to Recurring Decimal
*/
// @lc code=start
class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
int64_t n = numerator, d = denominator;
if (n == 0) return "0";
string res;
if (n < 0 ^ d < 0) res += '-';
n = labs(n);
d = labs(d);
res += to_string(n / d);
if (n % d == 0) return res;
res += '.';
unordered_map<int, int> m;
for (auto r = n % d; r > 0; r %= d) {
if (m.count(r)) {
res.insert(m[r], 1, '(');
res += ')';
break;
}
m[r] = res.size();
r *= 10;
res += to_string(r / d);
}
return res;
}
};
// @lc code=end
class Solution {
public String fractionToDecimal(int numerator, int denominator) {
long num = numerator, den = denominator;
if (num == 0 || den == 0) {
return "0";
}
StringBuilder sb = new StringBuilder();
boolean neg = (num > 0 && den < 0) || (num < 0 && den > 0);
num = Math.abs(num);
den = Math.abs(den);
sb.append((neg ? "-" : "") + String.valueOf(num / den));
if (num % den == 0) {
return sb.toString();
}
sb.append(".");
Map<Long, Integer> map = new HashMap<>();
num %= den;
map.put(num, sb.length());
while (num != 0) {
num *= 10;
sb.append(num / den);
num %= den;
if (map.containsKey(num)) {
sb.insert(map.get(num), "(");
sb.append(")");
break;
}
map.put(num, sb.length());
}
return sb.toString();
}
}
时间复杂度未知.