[LeetCode]Permutation Sequence

题目

Number: 60

Difficulty: Medium

Tags: Backtracking, Math

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.

题解

求解

n

个数的全排列中第

k

个排列。

对于

n

位数的全排列，如果第一个数字是

1

的排列有

(n-1)!

个，同样可得出每一位数字。

代码

[code]string getPermutation(int n, int k) {
    string res(n, '1');
    vector<int> l;
    for(int i = 0; i < n; i ++){
        l.push_back(i);
        res[i] = '1' + i;
    }
    if(k == 1)
        return res;
    vector<int> fac(n, 1);
    for(int i = n - 3; i >= 0; i--){
        fac[i] = fac[i + 1] * (n -i - 1);
    }
    k = k - 1;
    for(int i = 0; i < n; i++){
        int select = k / fac[i];
        k %= fac[i];
        res[i] = l[select ] + '1';
        remove(l.begin(), l.end(), l[select]);
    }
    return res;
}