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 { //本题思路:第n位确定时,会有(n-1)!个不同的排列组合。因此需要得到n!, //1.本题用count表示,每次循环时,count递减 //2.需要构造一个LinkedList链表,之所以不使用数组而使用链表,是因为删除链表的一个节点时,链表会自动补充之前的空缺位, //如果是数组,每次remove都需要重新排列。 //3.有一个细节,求第k位,首先k要减1,再进行计算 //一、康托展开:全排列到一个自然数的双射 //X=an*(n-1)!+an-1*(n-2)!+...+ai*(i-1)!+...+a2*1!+a1*0! //ai为整数,并且0<=ai<i(1<=i<=n) //适用范围:没有重复元素的全排列 public String getPermutation(int n, int k) { StringBuilder seq=new StringBuilder(); List<Integer> list=new LinkedList<Integer>(); int count=1; for(int i=0;i<n;i++){ list.add(i+1); count=count*(i+1); } k--;///// for(int i=0;i<n;i++){ count=count/(n-i); int index=k/count; seq.append(list.get(index)); list.remove(index); k=k%count; } return seq.toString(); } }
时间: 2024-10-08 00:49:49