Leetcode 60： Permutation Sequence

最新推荐文章于 2019-12-19 09:47:00 发布

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本文介绍了一种算法，用于解决给定整数n和k，返回第k个排列序列的问题。通过列举和标记所有从1到n的整数排列，可以得到n!个独特的排列。算法使用递减计数和阶乘计算的方法，逐步确定每个位置上的数字，最终形成所求的第k个排列。

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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 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.
Given k will be between 1 and n! inclusive.
Example 1:

Input: n = 3, k = 3
Output: “213”
Example 2:

Input: n = 4, k = 9
Output: “2314”

C++
string getPermutation(int n, int k) {
        vector<int> f(n+1,1);
        string num = "123456789";
        for(int i = 1;i < n+1;++i)
            f[i] = f[i-1]*i;
        string res = "";
        --k;
        for(int i = n;i > 0;--i)
        {
            int j = k/f[i-1];
            k = k%f[i-1];
            res.push_back(num[j]);
            num.erase(j,1);
        }
        return res;
    }