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