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LeetCode——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):
"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.
原题链接:https://oj.leetcode.com/problems/permutation-sequence/
从n个数的全排列中找出第k个排列。
n开头的排列有(n-1)!个。k/(n-1)!可确定第一个数字,在余下的(n-1)!中找k%(n-1)!个。
public class PermutationSequence { public String getPermutation(int n, int k) { List<Integer> list = new ArrayList<Integer>(); for(int i=1;i<=n;i++) list.add(i); int mod = 1; for (int i = 1; i <= n; i++) { mod = mod * i; } k--; StringBuilder builder = new StringBuilder(); for(int i=0;i<n;i++){ mod /= (n-i); int index = k / mod; k %= mod; builder.append(list.get(index)); list.remove(index); } return builder.toString(); } }
LeetCode——Permutation Sequence
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