166. fractions to decimals

166. Fractions to decimals

Original title link ：https://leetcode.cn/problems/fraction-to-recurring-decimal/

Given two integers , Molecules representing fractions respectively numerator Denominator denominator, With Returns a decimal as a string .

If the decimal part is a circular decimal , The loop is enclosed in brackets .

If there are multiple answers , Just go back Any one .

For all given inputs , Guarantee The length of the answer string is less than 104 .

Example 1：

Input ：numerator = 1, denominator = 2
Output ：“0.5”
Example 2：

Input ：numerator = 2, denominator = 1
Output ：“2”
Example 3：

Input ：numerator = 4, denominator = 333
Output ：“0.(012)”

Tips ：

-231 <= numerator, denominator <= 231 - 1
denominator != 0

Their thinking ：

First determine whether the denominator is 0, In judging the sign , Then determine whether the decimal number is circular . Use a dictionary to determine if decimals are circular ,key Is the remainder ,value Is the quotient of the remainder divided by the divisor , That is to return the corresponding index position of the number in the string .

Code implementation ：

class Solution:
    def fractionToDecimal(self, numerator: int, denominator: int) -> str:
        #  The denominator is 0
        if numerator == 0: 
            return "0"

        ans = []
        if (numerator > 0) ^ (denominator > 0):
            ans.append("-")

        numerator, denominator = abs(numerator), abs(denominator)
        a, b = divmod(numerator, denominator)
        ans.append(str(a))
        #  Remainder is 0, To divide or divide 
        if b == 0:
            return "".join(ans)
        #  There are decimals without division , Add a decimal point 
        ans.append(".")
        #  Use a dictionary to record where decimals begin 
        # key Is the remainder ,value The quotient of the remainder after division corresponds to ans Position in 
        loc = {
    b: len(ans)}
        #  Deal with the remainder 
        while b:
            #  The remainder is not enough for division , Expand the remainder by one digit , That's multiplied by 10
            b *= 10
            # a It's business. ,b Is the remainder , Did a division to update the remainder 
            a, b = divmod(b, denominator)
            ans.append(str(a))

            #  If the remainder appears in the dictionary , It means that the cycle begins to repeat 
            #  Add brackets according to the meaning of the question 
            if b in loc:
                ans.insert(loc[b], "(")
                ans.append(")")
                break
            #  Record the new remainder, and the quotient after division corresponds to ans Position in 
            loc[b] = len(ans)

        return "".join(ans)

reference ：
https://leetcode.cn/problems/fraction-to-recurring-decimal/solution/ji-lu-yu-shu-by-powcai/