[LeetCode] Fraction to Recurring Decimal 分數轉循環小數

 

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.

 

這道題仍是比較有意思的，開始還擔憂萬一結果是無限不循環小數怎麼辦，百度以後才發現原來能夠寫成分數的都是有理數，而有理數要麼是有限的，要麼是無限循環小數，無限不循環的叫無理數，例如圓周率pi或天然數e等，小學數學沒學好，汗！因爲還存在正負狀況，處理方式是按正數處理，符號最後在判斷，那麼咱們須要把除數和被除數取絕對值，那麼問題就來了：因爲整型數INT的取值範圍是-2147483648～2147483647，而對-2147483648取絕對值就會超出範圍，因此咱們須要先轉爲long long型再取絕對值。那麼怎麼樣找循環呢，確定是再獲得一個數字後要看看以前有沒有出現這個數。爲了節省搜索時間，咱們採用哈希表來存數每一個小數位上的數字。還有一個小技巧，因爲咱們要算出小數每一位，採起的方法是每次把餘數乘10，再除以除數，獲得的商即爲小數的下一位數字。等到新算出來的數字在以前出現過，則在循環開始出加左括號，結束處加右括號。代碼以下：

 

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<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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