[leetcode] Permutation Sequence

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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 k^th permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

第一个数的规律就是s[ (k-1)/(n-1)! ]

k的变化为 k = (k-1) % Factorial(n-1) + 1

好多k - 1很不爽，所以直接先k--，再做下面的

我第一次写居然用了递归@@ 弱爆了，直接一个while就可以了

class Solution {public:    int Factorial(int x) {        return (x == 1 ? x : x * Factorial(x - 1));    }    string getPermutation(int n, int k) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        string s(n,'1');        for(int i = 1; i <= n; i++){            s[i-1] = '0' + i;        }                k--;        //if(n == 1 || k == 0) return s;                string ret;        int idx;        while(s.size() > 1){            n = s.size();            idx = k / Factorial(n-1);            ret.push_back(s[idx]);            s.erase(s.begin() + idx);            k = k % Factorial(n-1);                    }                ret.push_back(s[0]);                return ret;            }};