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.
class Solution { public: string getPermutation(int n, int k) { string res; func(res,n,k); return res; } private: void func(string &res,int n,int k){ //n表示几位数 if(n==0) return; int weight = f(n-1); char cur = ‘1‘; int num = (k-1)/weight + 1; while( find(res.begin(),res.end(),cur)==res.end() || num!=0){ if(find(res.begin(),res.end(),cur)==res.end()) num--; if( num != 0) cur += 1; else break; } k %= weight; if(k==0) k = weight; res.push_back(cur); func(res,n-1,k); } int f(int n){ if(n == 1 || n==0) return 1; else return n*f(n-1); } };
[LeetCode] Permutation Sequence,布布扣,bubuko.com
[LeetCode] Permutation Sequence
原文：http://www.cnblogs.com/Xylophone/p/3920657.html