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)".

Solution:

    0.16  
6 ) 1.00
    0 
    1 0       <-- Remainder=1, mark 1 as seen at position=0.
    - 6 
      40      <-- Remainder=4, mark 4 as seen at position=1.
    - 36 
       4      <-- Remainder=4 was seen before at position=1, 
so the fractional part which is 16 starts repeating at position=1 => 1(6).

 

The key insight here is to notice that once the remainder starts repeating, so does the divided result.

You will need a hash table that maps from the remainder to its position of the fractional part. Once you found a repeating remainder, you may enclose the reoccurring fractional part with parentheses by consulting the position from the table.

The remainder could be zero while doing the division. That means there is no repeating fractional part and you should stop right away.

Just like the question Divide Two Integers, be wary of edge case such as negative fractions and nasty extreme case such as -2147483648 / -1.

public String fractionToDecimal(int numerator, int denominator) {
    if(denominator == 0) return "NaN"; //特殊情况1
    if(numerator == 0) return "0";  //特殊情况2
    // 如果两个数符号不同，结果为负数
    String sign = (denominator>>>31^numerator>>>31) == 1 ? "-" : "";
    
    // 先把除数和被除数转为long，以免除法和绝对值运算的时候溢出
    // 比如 -2147483648/-1 = -2147483648, abs(-2147483648) = -2147483648
    long num = numerator, den = denominator;
    num = Math.abs(num);
    den = Math.abs(den);
    long major = num / den;
    long rem = num % den;
    if(rem == 0) return sign+major;
    
    String pre = sign + major + ".";
    StringBuilder ans = new StringBuilder();
    Map<Long, Integer> map = new HashMap<Long, Integer>();
    while(rem != 0) { 
        int res = (int)(rem*10 / den);
        // 如果余数已经出现过一次，那么循环要开始了
        if(map.containsKey(rem)) {
            int index = map.get(rem);
            ans.insert(index, "(");
            ans.append(")");
            break;
        } else {
            ans.append(res);
            map.put(rem, ans.length()-1);
        }
        rem *= 10;
        rem %= den;
    }
    
    // 把整数部分和小数点插进去
    return ans.insert(0, pre).toString();
}

 

Solution2:

我们还可以通过LinkedHashSet来做。

public  String fractionToDecimal(int numerator, int denominator) {
    if(denominator == 0) return "NaN"; 
    if(numerator == 0) return "0";  
    String sign = (numerator>>>31^denominator>>>31) == 1 ? "-" : "";
    long a = numerator, b = denominator;
    a = Math.abs(a);
    b = Math.abs(b);
    String part1 = a / b + "";
    long mod = a % b;
    if(mod == 0) return sign+part1;
    long remain = mod;
    
    Set<Long> s = new LinkedHashSet<>();
    while (!s.contains(mod) && mod != 0) {
        s.add(mod);
        mod = (mod * 10) % b;
    }
    
    String part2 = "";
    boolean repeated = false;
    for (long i : s) {
        if (i == mod) {
            part2 += "(";
            repeated = true;
        }
        part2 += (remain * 10) / b;
        remain = (remain * 10) % b;
    }
    //  if (mod == 0)
    //      part2 += "(0";
    if(repeated) {
        part2 += ")";
    }
    
    return sign + part1 + "." + part2;
}