19.2.9 [LeetCode 60] Permutation Sequence

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:

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.
• 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"

题意

求出n个连续数的第k个permutation sequence

题解

class Solution {
public:
    string getPermutation(int n, int k) {
        if (n <= 1)return "1";
        vector<int>constn(n+1, 0);
        vector<char>nums(n + 1);
        int mult = 1;
        for (int i = 1; i <= n; i++) {
            mult *= i;
            constn[i] = mult;
            nums[i] = i + '0';
        }
        string ans = "";
        n--;
        int _n = n;
        while(n>1) {
            int idx;
            if(k%constn[n]==0)
                idx = k / constn[n];
            else
                idx = k / constn[n] + 1;
            ans += nums[idx];
            nums.erase(nums.begin() + idx);
            k = k % constn[n];
            n--;
            if (k == 0)break;
        }
        if (k == 1) {
            ans += nums[1];
            ans += nums[2];
        }
        else if (k == 2) {
            ans += nums[2];
            ans += nums[1];
        }
        else {
            for (int i = n+1; i >= 1; i--)
                ans += nums[i];
        }
        return ans;
    }
};

需要仔细一点