[Leetcode 60, Medium] Permutation Sequence

最新推荐文章于 2025-08-13 22:03:49 发布

simplify2012

最新推荐文章于 2025-08-13 22:03:49 发布

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分类专栏： Leetcode_medium 文章标签： leetcode

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本文探讨了如何通过数学分析和编程实现，找到一个给定范围内所有可能排列中指定位置的序列，具体步骤包括理解排列数量、构建候选列表并逐步筛选直至得到目标排列。

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Problem:

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.

Analysis:

Solutions:

C++:

    string getPermutation(int n, int k) {
        string rStr = "";
        if(n == 0 || k == 0)
            return rStr;
            
        if(n == 1)
            return "1";

        unsigned int num_of_perm = 1;
        for(int i = 1; i < n; ++i)
            num_of_perm *= i;

        vector<char> candidates;
        for(int i = 1; i <= n; ++i) {
            candidates.push_back(i + '0');
        }

        for(int i = 1; i < n && k != 0; ++i) {
            cout << "size = " << candidates.size() << ", k = " << k << endl;
            unsigned int group_num = (k - 1) / num_of_perm;
            rStr.push_back(candidates[group_num]);
            candidates.erase(candidates.begin() + group_num);

            k = k % num_of_perm;
            if(k == 0) {
                for(int j = candidates.size() - 1; j >= 0; --j)
                    rStr.push_back(candidates[j]);
                candidates.clear();
                break;
            }
            num_of_perm /= n - i;
        }

        return rStr;
    }

Java:

Python: