C++实现LeetCode(166.分数转循环小数)

• [LeetCode] 166.Fraction to Recurring Decimal 分数转循环小数

[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IN来源gaodai$ma#com搞$代*码*网T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：

 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 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; } };