1447. Simplified Fractions

Given an integer n, return a list of all simplified fractions between 0 and 1 (exclusive) such that the denominator is less-than-or-equal-to n. The fractions can be in any order.

Example 1:

Input: n = 2
Output: ["1/2"]
Explanation: "1/2" is the only unique fraction with a denominator less-than-or-equal-to 2.

Example 2:

Input: n = 3
Output: ["1/2","1/3","2/3"]

Example 3:

Input: n = 4
Output: ["1/2","1/3","1/4","2/3","3/4"]
Explanation: "2/4" is not a simplified fraction because it can be simplified to "1/2".

Example 4:

Input: n = 1
Output: []

Constraints:

• 1 <= n <= 100

class Solution {
    public List<String> simplifiedFractions(int n) {
        List<String> res = new ArrayList();
        if(n == 1) return new ArrayList();
        int k = n;
        while(k > 1){
            int t = k - 1; 
            while(t >= 1){
                if(gcd(k, t) != 1){
                t--;
                continue;
            }
                String sb = t + "/" + k;
                res.add(sb);
                t--;
            }
            k--;
        }
        return res;
    }
    public int gcd(int num1, int num2){
        int gcd = 1;
        for(int i = 1; i <= num1 && i <= num2; i++)
        {
            if(num1%i==0 && num2%i==0)
                gcd = i;
        }
        return gcd;
    }
}

狗急跳墙做法，遇到gcd不是1就continue，否则就添加到res中

class Solution {
    
    int gcd(int a, int b) {
        return b != 0 ? gcd(b, a % b) : a;
    }
    
    public List<String> simplifiedFractions(int n) {
        List<String> res = new ArrayList<>();
        for(int i = 1; i <= n; ++i) {
            for(int j = i + 1; j <= n; ++j) {
                if(gcd(i, j) > 1) continue;
                StringBuilder sb = new StringBuilder();
                sb.append(i).append('/').append(j);
                res.add(sb.toString());
            }
        }
        return res;
    }
}

简便写法

https://leetcode.com/problems/simplified-fractions/discuss/635685/Java-Clean-code-Use-GCD-w-Video-solution-explaining-intuition

扫描二维码关注公众号，回复： 11208031 查看本文章