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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.Subscribe to see which companies asked this question
思路
几个注意点： 正负号问题，numerator=0问题 解法是使用两个字典，一个字典key=余数，value=下一个相除结果 另外一个字典key=余数，value=当前的位置，同时计算时用一个列表将结果一一加入 余数==0或者余数已经在字典1中出现，则当前位置以前的序列是括号外面的字符串，当前位置以后是括号里面的字符串 注意一些细节，比如temp要重新赋值等等～～
class Solution(object): def fractionToDecimal(self,numerator,denominator): if (numerator>=0 and denominator>0) or (numerator<=0 and denominator<0): flag=1 else: flag=0 numerator=abs(numerator) denominator=abs(denominator) intenger=numerator/denominator remainder=numerator%denominator if remainder==0: if not flag: ss='-'+str(intenger) else: ss=str(intenger) return ss doc={} ans=[] id1={} i=0 while remainder!=0 and remainder not in doc: temp=remainder*10/denominator doc.setdefault(remainder,temp) id1.setdefault(remainder,i) i=i+1 ans.append(str(temp)) remainder=remainder*10%denominator temp=remainder*10/denominator if remainder: t1="".join([ans[i] for i in range(id1[remainder])]) t2="".join([ans[i] for i in range(id1[remainder],len(ans))]) if t1: s=str(intenger)+'.'+t1+'('+t2+')' else: s=str(intenger)+'.('+t2+')' else: ss="".join([ans[i] for i in range(len(ans))]) s=str(intenger)+'.'+ss if not flag: s='-'+s return s 转载地址：http://mwqmi.baihongyu.com/