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feat: add solutions to lc problem: No.0060
No.0060.Permutation Sequence
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‎solution/0000-0099/0060.Permutation Sequence/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举**
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我们知道，集合 $[1,2,..n]$ 一共有 $n!$ 种排列，如果我们确定首位，那剩余位能组成的排列数量为 $(n-1)!$。
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因此，我们枚举每一位 $i$，如果此时 $k$ 大于当前位置确定后的排列数量，那么我们可以直接减去这个数量；否则，说明我们找到了当前位置的数。
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对于每一位 $i$，其中 $0 \leq i \lt n$，剩余位能组成的排列数量为 $(n-i-1)!$，我们记为 $fact$。过程中已使用的数记录在 `vis` 中。
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时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def getPermutation(self, n: int, k: int) -> str:
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ans = []
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vis = [False] * (n + 1)
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for i in range(n):
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fact = 1
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for j in range(1, n - i):
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fact *= j
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for j in range(1, n + 1):
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if not vis[j]:
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if k > fact:
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k -= fact
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else:
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ans.append(str(j))
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vis[j] = True
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break
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return ''.join(ans)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String getPermutation(int n, int k) {
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StringBuilder ans = new StringBuilder();
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boolean[] vis = new boolean[n + 1];
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for (int i = 0; i < n; ++i) {
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int fact = 1;
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for (int j = 1; j < n - i; ++j) {
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fact *= j;
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}
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for (int j = 1; j <= n; ++j) {
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if (!vis[j]) {
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if (k > fact) {
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k -= fact;
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} else {
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ans.append(j);
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vis[j] = true;
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break;
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}
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}
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}
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}
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return ans.toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string getPermutation(int n, int k) {
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string ans;
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bitset<10> vis;
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for (int i = 0; i < n; ++i) {
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int fact = 1;
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for (int j = 1; j < n - i; ++j) fact *= j;
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for (int j = 1; j <= n; ++j) {
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if (vis[j]) continue;
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if (k > fact) k -= fact;
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else {
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ans += to_string(j);
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vis[j] = 1;
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break;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func getPermutation(n int, k int) string {
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ans := make([]byte, n)
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vis := make([]bool, n+1)
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for i := 0; i < n; i++ {
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fact := 1
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for j := 1; j < n-i; j++ {
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fact *= j
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}
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for j := 1; j <= n; j++ {
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if !vis[j] {
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if k > fact {
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k -= fact
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} else {
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ans[i] = byte('0' + j)
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vis[j] = true
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break
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}
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}
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}
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}
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return string(ans)
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}
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```
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### **C#**
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```cs
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public class Solution {
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public string GetPermutation(int n, int k) {
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var ans = new StringBuilder();
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int vis = 0;
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for (int i = 0; i < n; ++i) {
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int fact = 1;
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for (int j = 1; j < n - i; ++j) {
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fact *= j;
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}
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for (int j = 1; j <= n; ++j) {
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if (((vis >> j) & 1) == 0) {
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if (k > fact) {
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k -= fact;
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} else {
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ans.Append(j);
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vis |= 1 << j;
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break;
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}
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}
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}
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}
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return ans.ToString();
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}
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}
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```
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### **...**

‎solution/0000-0099/0060.Permutation Sequence/README_EN.md

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### **Python3**
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```python
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class Solution:
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def getPermutation(self, n: int, k: int) -> str:
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ans = []
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vis = [False] * (n + 1)
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for i in range(n):
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fact = 1
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for j in range(1, n - i):
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fact *= j
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for j in range(1, n + 1):
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if not vis[j]:
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if k > fact:
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k -= fact
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else:
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ans.append(str(j))
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vis[j] = True
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break
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return ''.join(ans)
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```
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### **Java**
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```java
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class Solution {
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public String getPermutation(int n, int k) {
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StringBuilder ans = new StringBuilder();
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boolean[] vis = new boolean[n + 1];
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for (int i = 0; i < n; ++i) {
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int fact = 1;
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for (int j = 1; j < n - i; ++j) {
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fact *= j;
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}
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for (int j = 1; j <= n; ++j) {
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if (!vis[j]) {
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if (k > fact) {
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k -= fact;
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} else {
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ans.append(j);
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vis[j] = true;
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break;
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}
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}
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}
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}
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return ans.toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string getPermutation(int n, int k) {
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string ans;
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bitset<10> vis;
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for (int i = 0; i < n; ++i) {
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int fact = 1;
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for (int j = 1; j < n - i; ++j) fact *= j;
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for (int j = 1; j <= n; ++j) {
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if (vis[j]) continue;
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if (k > fact) k -= fact;
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else {
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ans += to_string(j);
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vis[j] = 1;
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break;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func getPermutation(n int, k int) string {
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ans := make([]byte, n)
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vis := make([]bool, n+1)
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for i := 0; i < n; i++ {
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fact := 1
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for j := 1; j < n-i; j++ {
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fact *= j
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}
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for j := 1; j <= n; j++ {
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if !vis[j] {
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if k > fact {
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k -= fact
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} else {
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ans[i] = byte('0' + j)
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vis[j] = true
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break
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}
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}
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}
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}
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return string(ans)
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}
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```
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### **C#**
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```cs
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public class Solution {
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public string GetPermutation(int n, int k) {
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var ans = new StringBuilder();
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int vis = 0;
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for (int i = 0; i < n; ++i) {
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int fact = 1;
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for (int j = 1; j < n - i; ++j) {
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fact *= j;
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}
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for (int j = 1; j <= n; ++j) {
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if (((vis >> j) & 1) == 0) {
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if (k > fact) {
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k -= fact;
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} else {
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ans.Append(j);
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vis |= 1 << j;
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break;
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}
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}
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}
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}
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return ans.ToString();
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}
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}
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```
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### **...**

‎solution/0000-0099/0060.Permutation Sequence/Solution.cpp

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class Solution {
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public:
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string getPermutation(int n, int k) {
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string ans;
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bitset<10> vis;
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for (int i = 0; i < n; ++i) {
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int fact = 1;
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for (int j = 1; j < n - i; ++j) fact *= j;
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for (int j = 1; j <= n; ++j) {
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if (vis[j]) continue;
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if (k > fact) k -= fact;
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else {
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ans += to_string(j);
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vis[j] = 1;
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break;
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}
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}
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}
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return ans;
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}
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};

‎solution/0000-0099/0060.Permutation Sequence/Solution.cs

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using System.Collections.Generic;
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using System.Linq;
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using System.Text;
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public class Solution {
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public string GetPermutation(int n, int k) {
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--k;
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var factorial = Enumerable.Range(0, n).Select(i => Enumerable.Range(1, i).Aggregate(1, (agg, x) => agg * x)).ToArray();
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var numbers = new SortedSet<int>(Enumerable.Range(1, n));
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var sb = new StringBuilder();
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while (sb.Length < n)
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{
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var f = factorial[numbers.Count - 1];
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var index = k / f;
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var number = numbers.ElementAt(index);
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sb.Append(number);
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numbers.Remove(number);
18-
k %= f;
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var ans = new StringBuilder();
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int vis = 0;
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for (int i = 0; i < n; ++i) {
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int fact = 1;
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for (int j = 1; j < n - i; ++j) {
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fact *= j;
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}
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for (int j = 1; j <= n; ++j) {
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if (((vis >> j) & 1) == 0) {
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if (k > fact) {
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k -= fact;
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} else {
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ans.Append(j);
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vis |= 1 << j;
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break;
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}
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}
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}
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}
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return sb.ToString();
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return ans.ToString();
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}
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}
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}

‎solution/0000-0099/0060.Permutation Sequence/Solution.go

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Original file line numberDiff line numberDiff line change
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1+
func getPermutation(n int, k int) string {
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ans := make([]byte, n)
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vis := make([]bool, n+1)
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for i := 0; i < n; i++ {
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fact := 1
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for j := 1; j < n-i; j++ {
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fact *= j
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}
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for j := 1; j <= n; j++ {
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if !vis[j] {
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if k > fact {
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k -= fact
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} else {
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ans[i] = byte('0' + j)
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vis[j] = true
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break
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}
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}
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}
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}
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return string(ans)
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}

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