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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.
参考解析：http://bangbingsyb.blogspot.com/2014/11/leetcode-permutation-sequence.html
思路：先算n!。.然后k/(n-1)!＋1则为现在的第一个数字。更新k%=(n-1)!, n=n-1.
public class Solution { public String getPermutation(int n, int k) { if(n<=0) { return ""; } List<Integer> list=new ArrayList<Integer>(); int fac=1; for(int i=1;i<=n;i++) { list.add(i); fac*=i; } k--; //zero based StringBuilder sb=new StringBuilder(); while(n>0) { fac/=n; sb.append(list.remove(k/fac)); //di ji zu k%=fac; //di ji ge n--; } return sb.toString(); } }
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原文地址：http://www.cnblogs.com/Machelsky/p/5940776.html
