[LeetCode] 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)".

解题思路：

首先得到商的整数部分，余数部分不能整除则在余数后面加0。用一个vector保存每次相除后的余数，出现了相同的余数，则意味着有循环小数。

一个容易忽略的点是INT_MIN，即0x80000000，它的整数用int保存不了，需要转换格式为long long。

实现代码：

/*****************************************************************************
    *  @COPYRIGHT NOTICE
    *  @Copyright (c) 2015, 楚兴
    *  @All rights reserved
    *  @Version  : 1.0

    *  @Author   : 楚兴
    *  @Date     : 2015/2/6 19:57
    *  @Status   : Accepted
    *  @Runtime  : 18 ms
*****************************************************************************/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
	string fractionToDecimal(int numerator, int denominator) {
		if (denominator == 0)
		{
			return ""; //分母为0
		}
		if (numerator == 0)
		{
			return "0"; //分子为0
		}

		long long num = numerator;
		long long den = denominator;

		bool flag = false;
		if ((num > 0 && den < 0) ||(num < 0 && den > 0))
		{
			flag = true;
		}
		num = num >= 0 ? num : -num;
		den = den > 0 ? den : - den;

		string str = "";
		long long result = num / den; //商的整数部分
		char res[11];
		itoa(result,res);
		int len = strlen(res);
		str += res;

		long long n = num % den;
		if (!n)
		{
			if (flag)
			{
				str.insert(str.begin(),'-');
			}
			return str;
		}
		str += '.';
		
		vector<long long> tail;
		int index = -1;
		while(n)
		{
			tail.push_back(n);
			n *= 10;
			str += n / den + '0';
			n %= den;
			index = find(tail,n);
			if (index != -1)
			{
				break;
			}
		}
		if (index != -1)
		{
			str.insert(str.begin() + index + len + 1, '(');
			str.push_back(')');
		}
		if (flag)
		{
			str.insert(str.begin(),'-');
		}
		
		return str;
	}

	int find(vector<long long> num, long long n)
	{
		for (int i = 0; i < num.size(); i++)
		{
			if (num[i] == n)
			{
				return i;
			}
		}
		return -1;
	}

	void itoa(long long n, char* ch) //n不为负
	{
		char* temp = ch;
		if (n == 0)
		{
			*temp++ = '0';
			*temp = '\0';
			return;
		}
		while(n)
		{
			*temp++ = n % 10 + '0';
			n /= 10;
		}
		*temp = '\0';
		reverse(ch);
	}

	void reverse(char* ch)
	{
		char* start = ch;
		char* end = ch + strlen(ch) - 1;
		char c;
		while(start < end)
		{
			c = *start;
			*start++ = *end;
			*end-- = c;
		}
	}
};

int main()
{
	Solution s;
	cout<<s.fractionToDecimal(-2147483648, 1).c_str()<<endl;
	cout<<s.fractionToDecimal(-1,-2147483648).c_str()<<endl;
	cout<<s.fractionToDecimal(1,6).c_str()<<endl;
	cout<<s.fractionToDecimal(-6,-7).c_str()<<endl;
	system("pause");
}