leetcode | Permutation Sequence

Permutation Sequence : https://leetcode.com/problems/permutation-sequence/

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问题描述：

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.

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解析

1. 调用 next Permutation

调用上一题Next Permutation中的函数，逐步计算下一个队列，直到第k个。（暴力枚举）

    string getPermutation(int n, int k) {        string result;        vector<int> nums;        for (int i = 1; i <= n; i++)            nums.push_back(i);  //初始化第一个排列        for (int j = 1; j < k; j++)            nextPermutation(nums);  //计算第k个排列        for (int k = 0; k < n; k++)            result.push_back(nums[k]);         return result;    }

做了很多无用功，因为我们只需得到第 k 个排列，上述算法计算了所有排列，耗时太大，不能满足要求。

2. 数学解法

[1,2,3,…,n] 包含了 n! 种排列，那么以某一个数开头的排列有(n−1)!种， 如果 r=k / (n−1)!，则数字r位于第一位，然后从数据集中移出数字r 并且 k=k % (n−1)!，依次类推。
另外由
1. “123”
2. “132”
3. “213”
4. “231”
5. “312”
6. “321”可以观察到只有k−1才能保证k=1 123;k=2 132同时满足，第1位为1，即(2−1)/2!=0和(1−1)/2!=0

class Solution {public:    string getPermutation(int n, int k) {        string result;        vector<int> set;         for (int i = 1; i <= n; i++)            set.push_back(i);        k--;  //k-1后才能应用于整除        int count = factorial(n);  // n! 组合数        for (int j = n; j > 0; j--) {            count = count / j;  // (j-1)!            int r = k / count;            result.push_back(set[r]+'0');            set.erase(set.begin()+r);  //移除set[r]，或将r后的元素整体前移一位            k = k % count;        }        return result;    } private:    int factorial(int n) {  //求阶乘        if (n == 0)            return 1;        int result = 1;        while (n) {            result *= n;            n--;        }        return result;    }};