/* n个数有n!个排列,第k个排列,是以第(k-1)/(n-1)!个数开头的集合中第(k-1)%(n-1)!个数 */ public String getPermutation(int n, int k) { k--; List<Integer> list = new ArrayList<>(); StringBuilder res = new StringBuilder(); int count =1; //以每个数字开头的集合有多少中排列 for (int i = 2; i…
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" "…