166-Fraction to Recurring Decimal

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题意

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. If multiple answers are possible, return any of them. It is guaranteed that the length of the answer string is less than 104 for all the given inputs.

Example 1:
Input: numerator = 1, denominator = 2
Output: "0.5"

Example 2:
Input: numerator = 2, denominator = 1
Output: "2"

Example 3:
Input: numerator = 2, denominator = 3
Output: "0.(6)"

Example 4:
Input: numerator = 4, denominator = 333
Output: "0.(012)"

Example 5:
Input: numerator = 1, denominator = 5
Output: "0.2"

思路

• 数学题：特殊问题
• 注意小数点后面的算法：余数 * 10/ 除数
• 用HashMap记录 余数：字符串位置 -> 用于加’()’

class Solution(object):
    def fractionToDecimal(self, numerator, denominator):
        """
        :type numerator: int
        :type denominator: int
        :rtype: str
        """
        
        if numerator == 0: return '0'
        res = ''
        
        if (numerator < 0) ^ (denominator < 0): # 分子分母符号相异
            res += '-'

        num,den = abs(numerator),abs(denominator)
        res += str(num // den)
        if num % den == 0: return res
        else: res+='.'
        
        '''
        除尽：
        10 / 4 = 2...2    => 2.
                     num* 10 // den
                     2*10/4   => 5   余数不断 * 10 // 分母 -> num = 余数
                     
        除不尽
        10 /3 =3...1                  => res: 3.
                   num* 10/3
                   10 /3 ->3...1      => res: 3.3   map:{1:2}
                   1 in map           '3.' + '(' + '3' + ')'
                   
        
        '''
        num = num % den
        map = {num: len(res)} # map：记录当前余数：字符串位置
        while num:
            num *= 10
            res+= str(num//den)
            num %= den
            if num in map:
                res = res[:map[num]] +'(' + res[map[num]:] + ')'
                return res
            else:
                map[num] = len(res)

        return res

分析:

• Time: O(n)
• Space: O(n)