[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");}