【LeetCode】60. Permutation Sequence-优快云博客

本文介绍了一种高效算法，用于求解给定整数n和k时的第k个排列序列。通过数学规律和算法设计，实现了快速计算，适用于n在1到9之间的场景。

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题目描述

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):

"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.

解题思路

排列本身是有规律的，所以直接根据数学规律算出结果就好了。

AC代码

class Solution {
public:
    string getPermutation(int n, int k) {
        string ans;
        //记录数字1到n
        vector<char> nums;
        //记录1到n的阶乘，因为全排序的个数是n的阶乘
        vector<int> fac;
        fac.push_back(0);
        int mul = 1;
        for (int i = 1; i <= n; ++i) {
            nums.push_back(i + '0');
            mul *= i;
            fac.push_back(mul);
        }

        k -= 1;
        int curIdx = 0;
        for (int i = 0; i < n - 1; ++i) {
            curIdx = k / fac[n - i - 1];
            ans.push_back(nums[curIdx]);
            //当前数字已经加入排序中，移除
            nums.erase(nums.begin() + curIdx);
            k %= fac[n - i - 1];
        }
        ans.push_back(nums[0]);
        return ans;
    }
};