n0060_permutation_sequence.rs

/**
 * [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:
 *
 * <ol>
 * 	"123"
 * 	"132"
 * 	"213"
 * 	"231"
 * 	"312"
 * 	"321"
 * </ol>
 *
 * Given n and k, return the k^th 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"
 *
 *
 */
pub struct Solution {}

// submission codes start here

// target: split k = i! + j! + ...
use std::char::from_digit;
impl Solution {
    pub fn get_permutation(n: i32, k: i32) -> String {
        let factorials = [0, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880];
        let mut k = k;
        let mut i = n;
        let mut res = String::new();
        while i > 0 {
            if k > factorials[i as usize] {
                let round = k / factorials[i as usize];
                if round >= n {}
            } else {
                i -= 1;
            }
        }
        res
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_60() {}
}