题目
地址：https://leetcode.com/problems/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 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.
- 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"
回溯DFS实现
思路：
获取少一位的全排列，取商为相应的位置，把值取出来添加到结果StringBuilder对象。剩下余数继续上面的逻辑即可。
package backtracking;
import java.util.ArrayList;
import java.util.List;
// https://leetcode.com/problems/permutation-sequence/
public class PermutationSequence {
public static void main(String[] args) {
PermutationSequence obj = new PermutationSequence();
String result = obj.getPermutation(4, 9);
System.out.println("result > " + result);
}
public String getPermutation(int n, int k) {
// factorial
int[] factorial = new int[n + 1];
factorial[0] = 1;
int sum = 1;
// create an array of factorial lookup
for (int i = 1; i <= n; i++) {
sum *= i;
// factorial[] = {1, 1, 2, 6, 24, ..., n!}
factorial[i] = sum;
}
// create a list of numbers to get indices
List<Integer> items = new ArrayList<Integer>();
for (int i = 1; i <= n; i++) {
// numbers = {1, 2, 3, 4}
items.add(i);
}
// i from 0, so k - 1
k--;
StringBuilder sb = new StringBuilder();
// calculate
for (int i = 1; i <= n; i++) {
int index = k / factorial[n - i];
int item = items.get(index);
sb.append(String.valueOf(item));
items.remove(index);
k -= index * factorial[n - i];
}
return sb.toString();
}
}
英文详细分解
如果上面的代码看不明白，下面详细分解。
“Explain-like-I’m-five” Java Solution in O(n)
I’m sure somewhere can be simplified so it’d be nice if anyone can let me know. The pattern was that:
say n = 4, you have {1, 2, 3, 4}
If you were to list out all the permutations you have
1 + (permutations of 2, 3, 4)
2 + (permutations of 1, 3, 4)
3 + (permutations of 1, 2, 4)
4 + (permutations of 1, 2, 3)
We know how to calculate the number of permutations of n numbers… n! So each of those with permutations of 3 numbers means there are 6 possible permutations. Meaning there would be a total of 24 permutations in this particular one. So if you were to look for the (k = 14) 14th permutation, it would be in the
3 + (permutations of 1, 2, 4) subset.
To programmatically get that, you take k = 13 (subtract 1 because of things always starting at 0) and divide that by the 6 we got from the factorial, which would give you the index of the number you want. In the array {1, 2, 3, 4}, k/(n-1)! = 13/(4-1)! = 13/3! = 13/6 = 2. The array {1, 2, 3, 4} has a value of 3 at index 2. So the first number is a 3.
Then the problem repeats with less numbers.
The permutations of {1, 2, 4} would be:
1 + (permutations of 2, 4)
2 + (permutations of 1, 4)
4 + (permutations of 1, 2)
But our k is no longer the 14th, because in the previous step, we’ve already eliminated the 12. 4-number permutations starting with 1 and 2. So you subtract 12 from k… which gives you 1. Programmatically that would be…
k = k - (index from previous) * (n-1)! = k - 2*(n-1)! = 13 - 2*(3)! = 1
In this second step, permutations of 2 numbers has only 2 possibilities, meaning each of the three permutations listed above a has two possibilities, giving a total of 6. We’re looking for the first one, so that would be in the 1 + (permutations of 2, 4) subset.
Meaning: index to get number from is k / (n - 2)! = 1 / (4-2)! = 1 / 2! = 0… from {1, 2, 4}, index 0 is 1
so the numbers we have so far is 3, 1… and then repeating without explanations.
{2, 4}
k = k - (index from pervious) * (n-2)! = k - 0 * (n - 2)! = 1 - 0 = 1;
third number’s index = k / (n - 3)! = 1 / (4-3)! = 1/ 1! = 1… from {2, 4}, index 1 has 4
Third number is 4
{2}
k = k - (index from pervious) * (n - 3)! = k - 1 * (4 - 3)! = 1 - 1 = 0;
third number’s index = k / (n - 4)! = 0 / (4-4)! = 0/ 1 = 0… from {2}, index 0 has 2
Fourth number is 2
Giving us 3142. If you manually list out the permutations using DFS method, it would be 3142. Done! It really was all about pattern finding.
代码下载
参考
https://leetcode.com/problems/permutation-sequence/discuss/22507/%22Explain-like-I’m-five%22-Java-Solution-in-O(n)
