60. Permutation Sequence LeetCode] [k-th arrangement (Medium) (JAVA)
Topic Address: https://leetcode.com/problems/permutation-sequence/
Subject description:
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 for n = 3:
1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"
Given n and k, return the kth permutation sequence.
Note:
Given n will be between 1 and 9 inclusive.
Given k will be between 1 and n! inclusive.
Example 1:
Input: n = 3, k = 3
Output: "213"
Example 2:
Input: n = 4, k = 9
Output: "2314"
Subject to the effect
Given set [1,2,3, ..., n], a total of all the elements n! Permutations.
Problem-solving approach
From the k-th starting arrangement 1 is first started from 0 k - 1 Species
In fact, as long as the n-th element is calculated to
1, k / (n - 1 )! Can be calculated n-th element
2, the cycle of each element is calculated
class Solution {
public String getPermutation(int n, int k) {
if (n == 1 && k == 1) return "1";
List<Integer> list = new ArrayList<>();
int product = 1;
for (int i = 1; i <= n; i++) {
list.add(i);
if (i != n) product *= i;
}
k--;
StringBuilder res = new StringBuilder();
while (list.size() > 0) {
int cur = k / product;
res.append(list.remove(cur));
k = k % product;
product /= Math.max(1, list.size());
}
return res.toString();
}
}
When execution: 2 ms, beat the 81.25% of all users to submit in Java
memory consumption: 37.6 MB, beat the 5.78% of all users to submit in Java
