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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.
public class Solution { int f[]; public String getPermutation(int n, int k) { f = new int[n+1]; if(k>factorial(n)) return "-1"; List<Integer> list = new ArrayList<Integer>(); for(int i=1;i<=n;i++){ list.add(i); } StringBuilder sb = new StringBuilder(); k--; while(list.size()>0){ int mul = factorial(n-1); int index = k/mul; sb.append(list.get(index)); list.remove(index); k = k%mul; n--; } return sb.toString(); } private int factorial(int num){ if(num == 0) return f[0]=1; if(f[num]>0){ return f[num]; }else{ return f[num]=factorial(num-1)*num; } } }
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