QUESTION

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)".

1ST TRY

class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
char* ch = new char();
string ret;
string fractionStr = "";
int remain;
vector<bool> flag(, false); //integer part
*ch = numerator/denominator +'';
ret = ch; //fraction part
while()
{
remain = numerator%denominator;
if(remain == ) return ret;
else if(flag[remain]) return ret + ".(" + fractionStr + ")";
else
{
flag[remain] = true;
*ch = numerator/denominator +'';
fractionStr += ch;
}
}
}
};

Result: Runtime Error

Last executed input: -50, 8

2ND TRY

考虑了负数情况

class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
string intStr = "";
string fractionStr = "";
string tmpStr = "";
int remain;
vector<bool> flag(, false); // int 转 string
stringstream ss; //negative or not
if(numerator > && denominator < )
{
intStr = "-";
denominator = ~(denominator-);
}
else if(numerator < && denominator > )
{
intStr = "-";
numerator = ~(numerator-);
}
else if(numerator < && denominator < )
{
numerator = ~(numerator-);
denominator = ~(denominator-);
} //integer part
ss << numerator/denominator;
ss >> tmpStr;
intStr += tmpStr;
remain = numerator%denominator;
if(remain == ) return intStr; //fraction part
while()
{
if(remain == ) return intStr + "." + fractionStr;
else if(flag[remain]) return intStr + ".(" + fractionStr + ")";
else
{
flag[remain] = true;
numerator = remain * ;
remain = numerator%denominator;
ss.clear();
ss << numerator/denominator;
ss >> tmpStr;
fractionStr += tmpStr;
}
}
}
};

Result: Runtime Error

Last executed input: -2147483648, -10

3RD TRY

考虑溢出的情况

class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
long long int num = (long long int) numerator;
long long int den = (long long int) denominator; string intStr = "";
string fractionStr = "";
string tmpStr = "";
int remain;
set<int> flag; // int 转 string
stringstream ss; //negative or not
if(num > && denominator < )
{
intStr = "-";
den = ~(den-);
}
else if(num < && den > )
{
intStr = "-";
num = ~(num-);
}
else if(num < && den < )
{
num = ~(num-);
den = ~(den-);
} //integer part
ss << num/den;
ss >> tmpStr;
intStr += tmpStr;
remain = num%den;
if(remain == ) return intStr; //fraction part
while()
{
if(remain == ) return intStr + "." + fractionStr;
else if(flag.find(remain)!=flag.end()) return intStr + ".(" + fractionStr + ")";
else
{
flag.insert(remain);
num = remain * ;
remain = num%den;
ss.clear();
ss << num/den;
ss >> tmpStr;
fractionStr += tmpStr;
}
}
}
};

Result: Wrong

Input: 1, 6
Output: "0.(16)"
Expected: "0.1(6)"

4TH TRY

循环的位置得准确,所以用map代替set

class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
long long int num = (long long int) numerator;
long long int den = (long long int) denominator; string ret = "";
string tmpStr = "";
int remain;
map<int,int> flag; // int 转 string
stringstream ss; //negative or not
if(num > && denominator < )
{
ret = "-";
den = ~(den-);
}
else if(num < && den > )
{
ret = "-";
num = ~(num-);
}
else if(num < && den < )
{
num = ~(num-);
den = ~(den-);
} //integer part
ss << num/den;
ss >> tmpStr;
ret += tmpStr;
remain = num%den;
if(remain == ) return ret; //fraction part
ret += ".";
while()
{
if(remain == ) return ret;
else if(flag.find(remain)!=flag.end())
{
ret.insert(flag[remain], , '(');
return ret + ")";
}
else
{
flag[remain] = ret.length();
num = remain * ;
remain = num%den;
ss.clear();
ss << num/den;
ss >> tmpStr;
ret += tmpStr;
}
}
}
};

Result: Wrong

Input: -1, -2147483648
Output: "0.000000000000000000000000000000-1"
Expected: "0.0000000004656612873077392578125"

5TH TRY

remain也要申请为long long int, 否则在num = remain * 10;会溢出

class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
long long int num = (long long int) numerator;
long long int den = (long long int) denominator; string ret = "";
string tmpStr = "";
long long int remain;
map<int,int> flag; // int 转 string
stringstream ss; //negative or not
if(num > && denominator < )
{
ret = "-";
den = ~(den-);
}
else if(num < && den > )
{
ret = "-";
num = ~(num-);
}
else if(num < && den < )
{
num = ~(num-);
den = ~(den-);
} //integer part
ss << num/den;
ss >> tmpStr;
ret += tmpStr;
remain = num%den;
if(remain == ) return ret; //fraction part
ret += ".";
while()
{
if(remain == ) return ret;
else if(flag.find(remain)!=flag.end())
{
ret.insert(flag[remain], , '(');
return ret + ")";
}
else
{
flag[remain] = ret.length();
num = remain * ;
remain = num%den;
ss.clear();
ss << num/den;
ss >> tmpStr;
ret += tmpStr;
}
}
}
};

Result: Accepted

Fraction to Recurring Decimal(STRING-TYPE CONVERTION)的更多相关文章

  1. LeetCode解题报告—— Linked List Cycle II & Reverse Words in a String & Fraction to Recurring Decimal

    1. Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no ...

  2. 【LeetCode】166. Fraction to Recurring Decimal

    Fraction to Recurring Decimal Given two integers representing the numerator and denominator of a fra ...

  3. 【LeetCode】166. Fraction to Recurring Decimal 解题报告(Python)

    [LeetCode]166. Fraction to Recurring Decimal 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingz ...

  4. 【leetcode】Fraction to Recurring Decimal

    Fraction to Recurring Decimal Given two integers representing the numerator and denominator of a fra ...

  5. Leetcode 166. Fraction to Recurring Decimal 弗洛伊德判环

    分数转小数,要求输出循环小数 如2 3 输出0.(6) 弗洛伊德判环的原理是在一个圈里,如果一个人的速度是另一个人的两倍,那个人就能追上另一个人.代码中one就是速度1的人,而two就是速度为2的人. ...

  6. 【刷题-LeetCode】166 Fraction to Recurring Decimal

    Fraction to Recurring Decimal Given two integers representing the numerator and denominator of a fra ...

  7. LeetCode Fraction to Recurring Decimal

    原题链接在这里:https://leetcode.com/problems/fraction-to-recurring-decimal/ 题目: Given two integers represen ...

  8. 166. Fraction to Recurring Decimal -- 将除法的商表示成字符串(循环节用括号表示)

    Given two integers representing the numerator and denominator of a fraction, return the fraction in ...

  9. [LeetCode#116]Fraction to Recurring Decimal

    Problem: Given two integers representing the numerator and denominator of a fraction, return the fra ...

随机推荐

  1. APP-12-视觉技术-身份证识别

    1.Postman测试 图片转换为Base64:http://imgbase64.duoshitong.com/ Base64: Base64数据去掉表头文件:data:image/png;base6 ...

  2. ACM__01背包,完全背包,多重背包

    今天写题的时候碰到了一道完全背包题,可是没有看出来,乱写了一通,浪费了一个晚上,顺便复习一下背包的知识 01背包 每种物品只能选择一次或者不选,求背包容量内的最大价值 先给出状态转移方程: f[i][ ...

  3. U3D MemoryProfiler

    MemoryProfiler Unity 5.3a4 has a new very lowlevel memory profiler API. It can tell you which object ...

  4. 在系统中使用Bean Validation验证参数

    转自:http://www.importnew.com/18561.html 为什么要使用Bean Validation?  当我们实现某个接口时,都需要对入参数进行校验.例如下面的代码 1 2 3 ...

  5. 使用docker-compose创建启动容器时候提示服务名已存在

    今天在测试环境启动一个已有服务时候提示服务名已存在,造成这个原因是,前些天集群的一个节点挂了,导致这个节点上的docker容器全部没了,为了当时能正常使用就使用 docker-compose -H 1 ...

  6. Mybatis动态sql及性能优化-3

    内容简介 1.回顾 2.动态sql 3.性能优化 懒加载机制 一级缓存 二级缓存 一.回顾 1.config文件常用标签 properties标签:引入外部properties文件资源. settin ...

  7. 中文转码器的工作原理_delphi教程

    最近在做Delphi下的简体与繁体转换, 发现Windows2000自带的工具"中文转码器"很好用, 不仅可以转内码(BIG5-->GBK), 还可以将繁体字转为简体字(如: ...

  8. MyBatis对入参对象的属性空判断

      <!-- 查询学生list,like姓名 -->   <select id="getStudentListLikeName" parameterType=&q ...

  9. C++ 使用VS2010创建MFC ActiveX工程项目

    1.ActiveX的基本概念 ActiveX控件可以看作是一个极小的服务器应用程序,它不能独立运行,必须嵌入到某个容器程序中,与该容器一起运行.这个容器包括WEB网页,应用程序窗体等... Activ ...

  10. C# 图像处理:记录图像处理时间的一个类

    class HiPerTimer { [DllImport("user32.dll")] static extern bool GetLastInputInfo(ref LASTI ...