103:Fraction to Recurring Decimal
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题目: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)”.
解析:这题目难点在于如何找到无限循环的那一段。通过手工推演除法计算过程,我们会发现当一个余数第二次重复出现时,小数点后就开始循环了,代码如下:
// 通过手工计算除法,我们会发现当一个余数第二次重复出现时// 小数点后就开始无限循环了class Solution {public: string fractionToDecimal(int numerator, int denominator) { string res = ""; if (numerator == 0) return "0"; if (denominator == 0) return res; // int 表示的范围是 [-2147483648, 2147483647] // 因此 int 的最小值变为正数时会越界 // 所以把其赋给长整形 long long n = numerator; long d = denominator; if ((n < 0 && d > 0) || (n > 0 && d < 0)) res = "-"; if (n < 0) n = -n; if (d < 0) d = -d; res += to_string(n / d); n = n % d; if (n == 0) return res; res += '.'; int pos = res.size(); // map 用来记录被除数为 n 时相应的商在 // 返回字符串 res 中的位置,方便以后添加'(' unordered_map<long, int> record; // 模拟除法的过程 while (n != 0) { if (record.find(n) != record.end()) { res.insert(res.begin() + record[n], '('); res += ')'; return res; } record[n] = pos; // 表示 n * 10 / d 结果所插入的位置 res += char(n * 10 / d + '0'); ++pos; n = (n * 10) % d; } return res; }};
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