[LeetCode] 166.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.
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)".
Credits:
Special thanks to @Shangrila for adding this problem and creating all test cases.
这道题还是比较有意思的，开始还担心万一结果是无限不循环小数怎么办，百度之后才发现原来可以写成分数的都是有理数，而有理数要么是有限的，要么是无限循环小数，无限不循环的叫无理数，例如圆周率pi或自然数e等，小学数学没学好，汗！由于还存在正负情况，处理方式是按正数处理，符号最后在判断，那么我们需要把除数和被除数取绝对值，那么问题就来了：由于整型数INT的取值范围是-2147483648～2147483647，而对-2147483648取绝对值就会超出范围，所以我们需要先转为long long型再取绝对值。那么怎么样找循环呢，肯定是再得到一个数字后要看看之前有没有出现这个数。为了节省搜索时间，我们采用哈希表来存数每个小数位上的数字。还有一个小技巧，由于我们要算出小数每一位，采取的方法是每次把余数乘10，再除以除数，得到的商即为小数的下一位数字。等到新算出来的数字在之前出现过，则在循环开始出加左括号，结束处加右括号。代码如下：
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class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
int s1 = numerator >= 0 ? 1 : -1;
int s2 = denominator >= 0 ? 1 : -1;
long long num = abs( (long long)numerator );
long long den = abs( (long long)denominator );
long long out = num / den;
long long rem = num % den;
unordered_map<long long, int> m;
string res = to_string(out);
if (s1 * s2 == -1 && (out > 0 || rem > 0)) res = "-" + res;
if (rem == 0) return res;
res += ".";
string s = "";
int pos = 0;
while (rem != 0) {
if (m.find(rem) != m.end()) {
s.insert(m[rem], "(");
s += ")";
return res + s;
}
m[rem] = pos;
s += to_string((rem * 10) / den);
rem = (rem * 10) % den;
++pos;
}
return res + s;
}
};
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原文链接：https://www.cnblogs.com/grandyang/p/4238577.html