June Leetcoding Challenge(6.20)

LeetCode六月挑战（6.20 ）Permutation Sequence LeetCode 60

Solution

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.

Note:

• Given n will be between 1 and 9 inclusive.
• Given k will be between 1 and n! inclusive.

Example 1:

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

Example 2:

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

思路

代码

java

class Solution {
    public String getPermutation(int n, int k) {
        StringBuilder result = new StringBuilder(); 
        List<Integer> number = new ArrayList<>();
        int[] fib = new int[n];
        int jiec = 1;
        fib[0] = 1;
        for(int i=1;i<n;i++){
            fib[i] = i *fib[i-1];
        }
        
        for(int i=1;i<=n;i++)
             number.add(i);
        
        k=k-1;
        for(int i=n-1;i>=0;i--){
            int x = k/fib[i];
            System.out.println("x = "+x);
            result.append(number.get(x));
            k = k % fib[i];
            number.remove(x);
        }
        
        return result.toString();
    }
}