Simplified Fractions 之 判断最简分数

1.判断最简分数gcd函数

class Solution {
    public List<String> simplifiedFractions(int n) {
        List<String> res = new ArrayList<>();
        for(int molecule = 1; molecule < n; molecule++){
            for(int denominator = molecule+1; denominator <= n; denominator++){
                if(gcd(molecule, denominator) == 1){
                        String cur = molecule + "/" + denominator;
                        res.add(cur);
                }
            }
        }
        return res;
    }
    //判断是否为最简分数
    private int gcd(int x, int y){
        return x == 0 ? y : gcd(y % x, x);
    }
}

2.将分数转成另一个double变量，判断是否在set中。

class Solution {
    public List<String> simplifiedFractions(int n) {
        List<String> res = new ArrayList<>();
        Set<Double> set = new HashSet<>();
        for(int molecule = 1; molecule < n; molecule++){
            for(int denominator = molecule+1; denominator <= n; denominator++){
                if(set.add((double)molecule / denominator)){//判断是否为最简分数
                        String cur = molecule + "/" + denominator;
                        res.add(cur);
                }
            }
        }
        return res;
    }
}