LeetCode Permutation Sequence, permutationleetcode

LeetCode Permutation Sequence, permutationleetcode

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 ):

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.

Find the sequence of k.

Train of Thought: we can think of n numbers in a full arrangement of n !, For example, for a certain n, k, such as n = 4, k = 17. First, we know that there are 6, 2, and 1 combinations under n. So 17 = 2*6 + 2*2 + 1,

Because the weight of 6 is a multiple of 2 and there is surplus, the value of the 4th bits (highest bits) is a 1234 small number selected from "3rd", that is, 3;

Since the multiple of the Weight Value 2 is 2 and there is surplus, the value of the 3rd bits is to select a small number of 124 from "3rd", that is, 4;

Since the multiple of the weight value 1 is 1 and there is no surplus, the value of the 2nd bits is to select a small number of 1st from "12", that is, 1;

The last remaining number is appended in reverse order.

class Solution {public:    string remove(string s, int index, char &c) {        c = s[index-1];        string s1 = s.substr(0, index-1);        string s2 = s.substr(index);        return s1 + s2;    }        string getPermutation(int n, int k) {        if (n > 9 || n < 1) return "";                string temp = "123456789";        int A[9] = {1, 2, 6, 24, 120, 720, 5040, 40320, 362880};          if (A[n-1] < k) return "";                string s = temp.substr(0, n);        if (k == 1) return s;        if (k == A[n-1]) {            reverse(s.begin(), s.end());            return s;        }                string ans = "";        char c;        for (int i = n-2; i >= 0; i--) {            int m = 0;            if (k < A[i]) {                s = remove(s, 1, c);                ans += c;                continue;            }            while (k >= A[i]) {                k -= A[i];                m++;            }            if (k == 0) {                s = remove(s, m, c);                ans += c;                reverse(s.begin(), s.end());                return ans + s;            } else {                s = remove(s, m+1, c);                ans += c;            }        }                return ans;    }};