leetcode第一题怎么做（LeetCode力扣官方题解）
题目描述
给你一个整数 n，请你返回所有 0 到 1 之间（不包括 0 和 1）满足分母小于等于 n 的 最简 分数。分数可以以 任意 顺序返回。
示例 1：
输入：n = 2
输出：["1/2"]
解释："1/2" 是唯一一个分母小于等于 2 的最简分数。
示例 2：
输入：n = 3
输出：["1/2","1/3","2/3"]
示例 3：
输入：n = 4
输出：["1/2","1/3","1/4","2/3","3/4"]
解释："2/4" 不是最简分数，因为它可以化简为 "1/2" 。
示例 4：
输入：n = 1
输出：[]
提示：
- 1 <= n <= 100
由于要保证分数在（0，1）范围内，我们可以枚举分母 denominator ∈ [2,n] 和分子 numerator ∈ [1,denominator]，若分子分母的最大公约数为 1,则我们找到了一个最简分数。
代码
Python3
class Solution:
def simplifiedFractions(self, n: int) -> List[str]:
return [f"{numerator}/{denominator}" for denominator in range(2, n 1) for numerator in range(1, denominator) if gcd(denominator, numerator) == 1]
C
class Solution {
public:
vector<string> simplifiedFractions(int n) {
vector<string> ans;
for (int denominator = 2; denominator <= n; denominator) {
for (int numerator = 1; numerator < denominator; numerator) {
if (__gcd(numerator, denominator) == 1) {
ans.emplace_back(to_string(numerator) "/" to_string(denominator));
}
}
}
return ans;
}
};
Java
class Solution {
public List<String> simplifiedFractions(int n) {
List<String> ans = new ArrayList<String>();
for (int denominator = 2; denominator <= n; denominator) {
for (int numerator = 1; numerator < denominator; numerator) {
if (gcd(numerator, denominator) == 1) {
ans.add(numerator "/" denominator);
}
}
}
return ans;
}
public int gcd(int a, int b) {
return b != 0 ? gcd(b, a % b) : a;
}
}
C#
public class Solution {
public IList<string> SimplifiedFractions(int n) {
IList<string> ans = new List<string>();
for (int denominator = 2; denominator <= n; denominator) {
for (int numerator = 1; numerator < denominator; numerator) {
if (GCD(numerator, denominator) == 1) {
ans.Add(numerator "/" denominator);
}
}
}
return ans;
}
public int GCD(int a, int b) {
return b != 0 ? GCD(b, a % b) : a;
}
}
Golang
func simplifiedFractions(n int) (ans []string) {
for denominator := 2; denominator <= n; denominator {
for numerator := 1; numerator < denominator; numerator {
if gcd(numerator, denominator) == 1 {
ans = append(ans, strconv.Itoa(numerator) "/" strconv.Itoa(denominator))
}
}
}
return
}
func gcd(a, b int) int {
for a != 0 {
a, b = b%a, a
}
return b
}
C
#define MAX_FRACTION_LEN 10
int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
char ** simplifiedFractions(int n, int* returnSize) {
char ** ans = (char **)malloc(sizeof(char *) * n * (n - 1) / 2 );
int pos = 0;
for (int denominator = 2; denominator <= n; denominator ) {
for (int numerator = 1; numerator < denominator; numerator ) {
if (gcd(numerator, denominator) == 1) {
ans[pos] = (char *)malloc(sizeof(char) * MAX_FRACTION_LEN);
snprintf(ans[pos ], MAX_FRACTION_LEN, "%d%c%d", numerator, '/', denominator);
}
}
}
*returnSize = pos;
return ans;
}
JavaScript
var simplifiedFractions = function(n) {
const ans = [];
for (let denominator = 2; denominator <= n; denominator) {
for (let numerator = 1; numerator < denominator; numerator) {
if (gcd(numerator, denominator) == 1) {
ans.push(numerator "/" denominator);
}
}
}
return ans;
};
const gcd = (a, b) => {
if (b === 0) {
return a;
}
return gcd(b, a % b);
}
复杂度分析
- 时间复杂度: O(n2logn)。需要枚举 O(n2) 对分子分母的组合，每对分子分母计算最大公因数和生成字符串的复杂度均为 O(logn)。
- 空间复杂度:O(1) 。除答案数组外，我们只需要常数个变量。
BY /
本文作者：力扣
声明：本文归“力扣”版权所有，如需转载请联系。
,免责声明：本文仅代表文章作者的个人观点，与本站无关。其原创性、真实性以及文中陈述文字和内容未经本站证实，对本文以及其中全部或者部分内容文字的真实性、完整性和原创性本站不作任何保证或承诺，请读者仅作参考，并自行核实相关内容。文章投诉邮箱：anhduc.ph@yahoo.com
