【原创】leetCodeOj --- Fraction to Recurring Decimal 解题报告

原题地址：

https://oj.leetcode.com/problems/fraction-to-recurring-decimal/

题目内容：

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)".

方法：

要点有几个

0、模拟除法（自己做一遍，就是笔算复杂除法的过程。）

1、记住你产生的新的被除数以及其index，可以就用一个map来记，为了直观些我加了个set。

2、对产生的新被除数进行判断，如果重复了（map中有了），说明是循环小数，结束循环，在index前面加个括号，当然尾巴括号也要补上。

3、小心负数，由于正负不对称，为了方便我引入了long long类型，这样就不用担心溢出和负数了。

全部代码：

class Solution {
public:
    string fractionToDecimal(int numerator1, int denominator1) {
        bool isNegti = false;
	    long long numerator = numerator1;
	    long long denominator = denominator1;
        long long interger = numerator / denominator;
        if (numerator * denominator < 0) {
            isNegti  = true;
            interger = -interger;
        }
        numerator = numerator < 0 ? -numerator : numerator;
        denominator = denominator < 0 ? -denominator :denominator;
        long long next;
        if ((next = numerator % denominator) == 0)
            return isNegti == true ? "-" + strval(interger) : strval(interger);
        string deci  = ".";
        unordered_set<long long> dict;
        unordered_map<long long,long long> start;
        unordered_set<long long>::iterator itDict;
        unordered_map<long long,long long>::iterator itStart;
        char nownum;
        int index = 1;
        while (next) {
            itDict = dict.find(next);
            if (itDict != dict.end()) {
                itStart = start.find(next);
                int len = itStart->second;
                deci = deci.substr(0,len) + '(' + deci.substr(len) + ')';
                break;
            }
            dict.insert(next);
            start[next] = index ++;
            nownum = (next * 10) / denominator + '0';
            deci = deci + nownum;
            next = (next * 10) % denominator;
        }
        return isNegti == true ? "-" + strval(interger) + deci : strval(interger) + deci;
    }

    static string strval(long long num) {
        char tmp[20];
        char index = 0;
        while (num >= 10) {
            tmp[index ++] = (char)(num % 10 + '0');
            num /= 10;
        }
        tmp[index] = (char)(num + '0');
        string res = "";
        for (int i = index; i >= 0; i --) {
            res = res + tmp[i];
        }
        return res;
    }
};