LeetCode 166. Fraction to Recurring Decimal

最新推荐文章于 2022-10-01 23:11:08 发布

转载最新推荐文章于 2022-10-01 23:11:08 发布 · 298 阅读

LeetCode 题解同时被 2 个专栏收录

288 篇文章

订阅专栏

Multiple Passes

85 篇文章

订阅专栏

本文介绍了一种使用C++实现将任意分数转换为十进制小数的方法，并特别关注了循环小数的处理过程。通过记录余数的位置来标识循环开始的位置，从而在输出的小数中正确插入括号表示循环节。

So, the best solution I have seen. 点击打开链接

Integer part is easy to handle. For the fraction part, we need to remember where the loop start. Thus, we need to remember the position to insert '(' and the remainder.

#include <string>
#include <unordered_map>
#include <iostream>
using namespace std;

string getDec(long remainder, long den) {
  string res = "";
  unordered_map<long, int> map;
  int i = 0;
  while((remainder != 0) && (map.find(remainder) == map.end())) {
    map.insert(make_pair(remainder, i));
    i++;
    remainder = remainder * 10;
    res = res + (to_string(remainder/den));
    remainder = remainder % den;
  }
  if(remainder != 0) {
    int pos = map[remainder];
    res.insert(pos, 1, '('); // insert position, bytes, char
    res = res + ")";
  }
  return res;
}

string fractionToDecimal(int numerator, int demoinator) {
  long num = numerator;
  long den = demoinator;

  bool neg = num * den < 0;
  num = abs(num);
  den = abs(den);

  string res = neg ? ("-" + to_string(num / den)) : to_string(num / den);
  long remainder = num % den;

  return (remainder == 0) ? res : (res + "." + getDec(remainder, den));
}

int main(void) {
  cout << fractionToDecimal(4, 333) << endl;
}