[leetcode] 166. Fraction to Recurring Decimal 解题报告

题目链接: https://leetcode.com/problems/fraction-to-recurring-decimal/

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)".

Hint:

1. No scary math, just apply elementary math knowledge. Still remember how to perform a long division?
2. Try a long division on 4/9, the repeating part is obvious. Now try 4/333. Do you see a pattern?
3. Be wary of edge cases! List out as many test cases as you can think of and test your code thoroughly.

思路: 有以下几种情况:

1. 如果可以直接除尽, 那么最好, 直接返回

2. 如果不能直接除尽, 那么先取整数, 再计算小数. 计算小数部分时就是每次取商, 又可分为能除尽, 和循环小数

       1. 如果能除尽那么当最后可以整除的时候返回结果

       2. 比较麻烦的是循环小数. 我们需要将每次的被除数和当前商的位置用一个hash表存起来, 这样当某次发现相同的

           被除数时说明出现了循环, 那么就加在第一次出现这个被除数的商的位置加括号

并且在做运算的时候可能会越界, 因此我们要将除数和被除数都以long类型存储, 并且在运算之前将符号先提取出来.

写的我头好昏! ^.^

代码如下: 

class Solution {
public:
    string fractionToDecimal(int numerator, int denominator) {
        if(numerator == 0) return "0";
        long m=labs(numerator), n=labs(denominator);
        unordered_map<int, int> hash;
        int k=numerator>0 ^ denominator>0, i = 0; 
        string result;
        if(m % n == 0) return k==0?to_string(m/n):"-"+to_string(m/n);
        result = to_string(m/n) + ".";
        i = result.size();
        m = m % n * 10;
        while(hash.find(m) == hash.end())
        {
            hash[m] = i++;
            if(m % n ==0)
            {
                result += to_string(m/n);
                return k==0?result:"-"+result;
            }
            result += to_string(m/n);
            m = m%n*10;
        }
        result.insert(hash[m], 1, '(');
        result+= ")";
        return k==0?result:"-"+result;
    }
};