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):
"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.
Solution：
思路：根据排列组合数量 找位置
Time Complexity: O(N) Space Complexity: O(N)
Solution Code：
public class Solution {
public String getPermutation(int n, int k) {
int pos = 0;
List<Integer> numbers = new ArrayList<>();
int[] factorial = new int[n+1];
StringBuilder sb = new StringBuilder();
// create an array of factorial lookup
int sum = 1;
factorial[0] = 1;
for(int i=1; i<=n; i++){
sum *= i;
factorial[i] = sum;
}
// factorial[] = {1, 1, 2, 6, 24, ... n!}
// create a list of numbers to get indices
for(int i=1; i<=n; i++){
numbers.add(i);
}
// numbers = {1, 2, 3, 4}
k--;
for(int i = 1; i <= n; i++){
int index = k/factorial[n-i];
sb.append(String.valueOf(numbers.get(index)));
numbers.remove(index);
k-=index*factorial[n-i];
}
return String.valueOf(sb);
}
}
