leetcode-每日一题2022.2.10 最简分数

题目

力扣

思路 暴力

遍历分子和分母，判断最小公因数是1的话，就添加进结果中。

求最小公因素的笨蛋代码

class Solution {
public:
    vector<string> simplifiedFractions(int n) {
        vector<string> ans;
        for(int i = 1; i < n; i++){
            for(int j = i+1; j <= n; j++){
                bool flag = true;
                for(int k = 2; k <= i; k++){
                    if(i % k == 0 && j % k ==0){
                        // cout<<i<<' '<<j<<endl;
                        flag = false;
                        break;
                    }
                }
                if(!flag) continue;
                ans.push_back(to_string(i)+'/'+to_string(j));
            }
        }
        return ans;
    }
};

C++内置函数__gcd

class Solution {
public:
    vector<string> simplifiedFractions(int n) {
        vector<string> ans;
        for(int i = 1; i < n; i++){
            for(int j = i+1; j <= n; j++){
                if(__gcd(i,j)==1)
                    ans.push_back(to_string(i)+'/'+to_string(j));
            }
        }
        return ans;
    }
};

欧几里得算法

class Solution {
public:
    int gcd(int a,int b){
        return b == 0 ? a : gcd(b, a % b);
    }
    vector<string> simplifiedFractions(int n) {
        vector<string> ans;
        for(int i = 1; i < n; i++){
            for(int j = i+1; j <= n; j++){
                if(gcd(i,j)==1)
                    ans.push_back(to_string(i)+'/'+to_string(j));
            }
        }
        return ans;
    }
};

标签：10,gcd,简分数,int,public,vector,ans,2022.2,class
来源： https://blog.csdn.net/ZHUYAN1209/article/details/122871642