Leetcode-166.Fraction to Recurring Decimal

解析分数转换为小数并处理循环节
本文介绍了一种方法将两个整数表示的分数转换为字符串格式的小数,并特别关注处理循环节的过程。通过使用哈希表记录余数,文章详细解释了如何在发现重复余数时正确地封闭循环节。

Problem Description:
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)”.


Analysis:
First, we need to know in what case that the decimal part of fraction will recur. Answer is remainders should recur. For example 4 / 300, the remainders are 4, 40, 100, 100….So we need to use a hash table to record the remainders. Once we find a repeated remainder, we know that we have reached the end of the recurring parts and should enclose it with a ) . Also we need to store the position of the corresponding quotient digit of each remainder in the recurring parts .
ps :
1. careful with the zero numerator.
2. pay attention to the sign
3. maybe overflow (r *= 10);
4. Three cases : no fractional part ; fractional part does not recur; fractional part recurs ;

code :

class Solution {
public:
    string fractionToDecimal(int numerator, int denominator) {
        long n = numerator, d = denominator;
         // zero numerator
        if (n == 0) return "0";
        string res;
        //determine the sign
        if (n < 0 ^ d < 0) res+= '-';
        // remove the sign of operands
        n = abs(n), d = abs(d);
        res += to_string(n / d);

        if(n % d == 0) return res; //case 1.
        res += '.';
        unordered_map<int, int> map;
        // simulate the division process
        for (long r = n % d; r; r %= d)
        {
            //cout<<r<<endl;
            //**meet a known remainder**
            // so we reach the end of the repeating part
            if (map.count(r) > 0)
            {
                res.insert(map[r], 1,'(');//  augments : pos ,size_t n, char c;
                res += ')';
                break;
            }
             // the remainder is first seen
        // **remember the current position for it**
            map[r] = res.size();
            cout<< " s:"<<map[r]<<endl;
            r *= 10;
            //append the quotient digit
             res += to_string(r / d);
        }
    return res;
    }
};
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