[LeetCode]Fraction to Recurring Decimal

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)".

分析

这道题不需要特别的算法。只是在算小数的时候，用个hashmap存余值，从而方便发现循环点。

这种除法的问题，写的时候想一遍过还是得注意很多情况。比如结果正负号，越界，输入为零等情况。我们可以在开始就判断结果的正负，然后直接算绝对值的除法。为避免越界，我们全部变量用Long代替，输入为零的情况也可以开始就先拿出考虑。

复杂度

time: O(n), space: O(n)

代码

public class Solution {
    public String fractionToDecimal(int numerator, int denominator) {
        // 先考虑输入为零的情况
        if (numerator == 0)
            return "0";
        if (denominator == 0)
            return "";
        
        StringBuilder sb = new StringBuilder();
        if ((numerator ^ denominator) >>> 31 == 1) { // 确定结果正负
            sb.append('-');
        }
        
        // 避免越界，变量类型全为Long
        long a = Math.abs((long)numerator);
        long b = Math.abs((long)denominator);
        long intVal = a / b;
        long remainder = a % b;
        sb.append(intVal); // 添加整数部分
        
        // 开始考虑小数部分
        if (remainder > 0) {
            sb.append('.');
            int i = sb.length();
            Map<Long, Integer>  map = new HashMap<>();
            while (remainder > 0) {
                if (map.containsKey(remainder)) { // 发现小数循环
                    int start = map.get(remainder);
                    sb.insert(start, "(");
                    sb.append(")");
                    break;
                }
                map.put(remainder, i++);
                a = remainder * 10;
                long val = a / b;
                sb.append(val);
                remainder = a % b;
            }
        }
        return sb.toString();
    }
}