2.1.13 Permutation Sequence

Link: https://oj.leetcode.com/problems/permutation-sequence/

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 (ie, 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.

這題完全沒有思路。再做。

以下代碼參考：http://www.lifeincode.net/programming/leetcode-permutation-sequence-java/

Time: O(n), 

But this blog said its O(n^2) because removing elements from array is O(n) //What if removing an element from an ArrayList?

• ArrayList 對於隨機位置的add/remove，時間複雜度爲 O(n),但是對於列表末尾的添加/刪除操作,時間複雜度是 O(1). (Ref: http://blog.csdn.net/renfufei/article/details/17077425)

http://blog.csdn.net/linhuanmars/article/details/22028697

public class Solution {
    public String getPermutation(int n, int k) {
        int t = 1;
        ArrayList<Integer> num = new ArrayList<Integer> ();
        for(int i = 1; i <=n; i++){
            t *= i;
            num.add(i);//num = {1, 2, 3}
        }
        //t = n!
        k--;//from 1-indexed to 0-indexed
        t /=n;//t = (n-1)!
        StringBuffer sb = new StringBuffer();
        for(int i = n-1; i >=1; i--){//i starts with n-1 because t = (n-1)!, need to divide (n-1)
            int p = k/t;
            int digit = num.get(p);
            sb.append(String.valueOf(digit));//note to use "String.valueOf"!
            num.remove(p);
            k %= t;
            t /=i;
        }
        sb.append(String.valueOf(num.get(0)));
        return sb.toString();
    }
}