60. Permutation Sequence

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:

“123”
“132”
“213”
“231”
“312”
“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”

Problem URL

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Solution

给一个1-9的数字n，找到n!的第k个数字，k的范围是1-n!。

We could find that n! is composed of n * (n - 1)! using this we could use a factors[n + 1] to store 0! to n! and a list number denotes 1 - 9 for easy remove. After initialize them, we use k to divide factors[n-1] which would help us get the first digit of kth number. Append it to a stringbuilder and remove it from numbers. Then k should minus index multiple factors[n - 1].

Code

class Solution {
    public String getPermutation(int n, int k) {
        int[] factors = new int[n + 1];
        List<Integer> numbers = new ArrayList<>();
        StringBuilder sb = new StringBuilder();
        //initialize factors
        factors[0] = 1;
        int sum = 1;
        for (int i = 1; i <= n; i++){
            sum *= i;
            factors[i] = sum;
        }
        //initialize numbers for get the answer;
        for (int i = 1; i <= n; i++){
            numbers.add(i);
        }
        //start find kth permuation, let k--.
        k--;        
        for (int i = 1; i <= n; i++){
            int index = k / factors[n - i];
            sb.append(String.valueOf(numbers.get(index)));
            numbers.remove(index);
            k -= index * factors[n - i];
        }
        
        return sb.toString();
    }
}

Time Complexity: O(nk)
Space Complexity: O(k)

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