LeetCode-60-Permuataion Sequence

最新推荐文章于 2024-08-30 22:58:17 发布

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最新推荐文章于 2024-08-30 22:58:17 发布

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原文链接：http://www.cnblogs.com/nobodywang/p/10337869.html

本文介绍了一种求解在1到n范围内第k个排列序列的算法。通过理解排列与阶乘的关系，利用递归思想，实现了从n个数中找到特定位置的数，并移除已使用数的过程。该算法适用于n在1到9之间的范围，k在1到n!之间。

算法描述：

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 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"

解题思路：理解排列与阶乘的关系。n个数的排列方式有 n! 种，而n-1个数的排列方式有 (n-1)! 种。所以通过 k 除以 (n-1)! 得到对应位置的索引值。注意去除用过的数。

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