Question

60. Permutation Sequence

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Description

The set [1, 2, 3, ..., n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order, we get the following sequence for n = 3:

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the kth permutation sequence.

 

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

Example 3:

Input: n = 3, k = 1
Output: "123"

 

Constraints:

• 1 <= n <= 9
• 1 <= k <= n!

Solutions

Solution 1: Enumeration

We know that the set $[1,2,..n]$ has a total of $n!$ permutations. If we determine the first digit, the number of permutations that the remaining digits can form is $(n-1)!$.

Therefore, we enumerate each digit $i$. If $k$ is greater than the number of permutations after the current position is determined, then we can directly subtract this number; otherwise, it means that we have found the number at the current position.

For each digit $i$, where $0 \leq i < n$, the number of permutations that the remaining digits can form is $(n-i-1)!$, which we denote as $fact$. The numbers used in the process are recorded in vis.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$.

class Solution:
  def getPermutation(self, n: int, k: int) -> str:
    ans = []
    vis = [False] * (n + 1)
    for i in range(n):
      fact = 1
      for j in range(1, n - i):
        fact *= j
      for j in range(1, n + 1):
        if not vis[j]:
          if k > fact:
            k -= fact
          else:
            ans.append(str(j))
            vis[j] = True
            break
    return ''.join(ans)

class Solution {
  public String getPermutation(int n, int k) {
    StringBuilder ans = new StringBuilder();
    boolean[] vis = new boolean[n + 1];
    for (int i = 0; i < n; ++i) {
      int fact = 1;
      for (int j = 1; j < n - i; ++j) {
        fact *= j;
      }
      for (int j = 1; j <= n; ++j) {
        if (!vis[j]) {
          if (k > fact) {
            k -= fact;
          } else {
            ans.append(j);
            vis[j] = true;
            break;
          }
        }
      }
    }
    return ans.toString();
  }
}

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

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

impl Solution {
  pub fn get_permutation(n: i32, k: i32) -> String {
    let mut k = k;
    let mut ans = String::new();
    let mut fact = vec![1; n as usize];
    for i in 1..n as usize {
      fact[i] = fact[i - 1] * (i as i32);
    }
    let mut vis = vec![false; n as usize + 1];

    for i in 0..n as usize {
      let cnt = fact[(n as usize) - i - 1];
      for j in 1..=n {
        if vis[j as usize] {
          continue;
        }
        if k > cnt {
          k -= cnt;
        } else {
          ans.push_str(&j.to_string());
          vis[j as usize] = true;
          break;
        }
      }
    }

    ans
  }
}

public class Solution {
  public string GetPermutation(int n, int k) {
    var ans = new StringBuilder();
    int vis = 0;
    for (int i = 0; i < n; ++i) {
      int fact = 1;
      for (int j = 1; j < n - i; ++j) {
        fact *= j;
      }
      for (int j = 1; j <= n; ++j) {
        if (((vis >> j) & 1) == 0) {
          if (k > fact) {
            k -= fact;
          } else {
            ans.Append(j);
            vis |= 1 << j;
            break;
          }
        }
      }
    }
    return ans.ToString();
  }
}