LeetCode 60. Permutation Sequence

1. 题目要求

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.

2. 解题思路

最朴素的思想是回溯， 不过这里可以比较讨巧进行计算得到，n - 1 个排列数总共有 (n - 1)! 种排列方式， 如果 k 大于 (n - 1)! 表明目标应该在 下一段 (n - 1)! 的排列序列中。

3. code

class Solution {public:    Solution(){        arr = vector<int>{ 1, 1, 2, 6, 24, 120, 720, 5040, 40320 };    }public:    string getPermutation(int n, int k) {        vector<int> myset;        for (int i = 0; i != n; i++)            myset.push_back(i + 1);        return getPermutation(n, k - 1, myset);    }private:    string getPermutation(int n, int k, vector<int> & myset) {        if (myset.size() == 1)            return to_string(myset[0]);        string res;        int cur = k / arr[n - 1];        int data = myset[cur];        myset.erase(myset.begin() + cur);        return to_string(data) + getPermutation(n - 1, k - arr[n - 1] * cur, myset);    }    vector<int> arr;};

4. 大神解法

类似的思想， 不过有详细的说明~~

/*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 have1 + (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 the3 + (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)! = 1In 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 1so 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 2Giving us 3142. If you manually list out the permutations using DFS method, it would be 3142. Done! It really was all about pattern finding.*/public class Solution {public String getPermutation(int n, int k) {    int pos = 0;    List<Integer> numbers = new ArrayList<>();    int[] factorial = new int[n+1];    StringBuilder sb = new StringBuilder();    // create an array of factorial lookup    int sum = 1;    factorial[0] = 1;    for(int i=1; i<=n; i++){        sum *= i;        factorial[i] = sum;    }    // factorial[] = {1, 1, 2, 6, 24, ... n!}    // create a list of numbers to get indices    for(int i=1; i<=n; i++){        numbers.add(i);    }    // numbers = {1, 2, 3, 4}    k--;    for(int i = 1; i <= n; i++){        int index = k/factorial[n-i];        sb.append(String.valueOf(numbers.get(index)));        numbers.remove(index);        k-=index*factorial[n-i];    }    return String.valueOf(sb);}}