60. Permutation Sequence（难）

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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.

参考Permutation Sequence -- LeetCode

思路是这样的，比如当前长度是n，我们知道每个相同的起始元素对应(n-1)!个permutation，也就是(n-1)!个permutation后会换一个起始元素。因此，只要当前的k进行(n-1)!取余，得到的数字就是当前剩余数组的index，如此就可以得到对应的元素。如此递推直到数组中没有元素结束。实现中我们要维护一个数组来记录当前的元素，每次得到一个元素加入结果数组，然后从剩余数组中移除。

class Solution {public:string getPermutation(int n, int k) {k--;vector<int> nums;for (int i = 1; i <= n; i++){nums.push_back(i);}int fac = 1;for (int i = 2; i < n; i++){fac *= i;}string res;int round = n - 1;while (round >= 0){int index = k / fac;k %= fac;res += nums[index] + '0';nums.erase(nums.begin()+index);if (round){fac /= round;}round--;}return res;}};

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