[LeetCode] Permutation Sequence

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.

Solution: From left to right, we compute each digit in the desired sequence.
Some simple observations:
1. If k denotes the last sequence, construct the last sequence and return;
2. Compute the highest digit, and update k. 

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public class Solution { public String getPermutation(int n, int k) { int[] factorial = new int[n]; factorial[0] = 1; for (int i = 1; i < n; i++) { factorial[i] = factorial[i - 1] * (i + 1); } ArrayList<Integer> list = new ArrayList<Integer>(); for (int i = 1; i <= n; i++) { list.add(i); } StringBuffer sb = new StringBuffer(); int m = n - 1; while (k > 0) { if (k == factorial[m]) { for (int i = list.size() - 1; i >= 0; i--) { sb.append(list.get(i)); } return sb.toString(); } else { int t = (k - 1) / factorial[m - 1]; sb.append(list.remove(t)); k = k - t * factorial[m - 1]; m--; } } return sb.toString(); } }