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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 kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
1 string getPermutation(int n, int k) 2 { 3 string s; 4 if (k <= 0 || n <= 0) 5 return s; 6 7 int *array = new int[n], *fact = new int[n + 1], select, i, j; 8 fact[0] = 1; 9 for (i = 0; i < n; i++) 10 { 11 array[i] = ‘1‘ + i; 12 fact[i + 1] = fact[i] * (i + 1); 13 } 14 15 for (i = 0; i < n; i++) 16 { 17 select = (k - 1) / fact[n - i - 1]; 18 k = k - select * fact[n - i - 1]; 19 s += array[select]; 20 for (j = select; j < n - 1; j++) 21 array[j] = array[j + 1]; 22 } 23 24 delete[] array; 25 return s; 26 }
leetcode. Permutation Sequence
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原文地址：http://www.cnblogs.com/ym65536/p/4190066.html
