carloscn
commented
1 year ago Analysis
pub fn construct2_d_array(original: Vec<i32>, m: i32, n: i32) -> Vec<Vec<i32>>
{
let mut ret:Vec<Vec<i32>> = vec![vec![]];
if original.len() < 1 || m * n != original.len() as i32 {
return ret;
}
ret = vec![vec![0; n as usize]; m as usize];
for i in 0..(m) as usize {
for j in 0..(n) as usize {
ret[i][j] = original[j + i*(n as usize)];
}
}
return ret;
}
Description
You are given a 0-indexed 1-dimensional (1D) integer array original, and two integers, m and n. You are tasked with creating a 2-dimensional (2D) array with m rows and n columns using all the elements from original.
The elements from indices 0 to n - 1 (inclusive) of original should form the first row of the constructed 2D array, the elements from indices n to 2 * n - 1 (inclusive) should form the second row of the constructed 2D array, and so on.
Return an m x n 2D array constructed according to the above procedure, or an empty 2D array if it is impossible.
Example 1:
Input: original = [1,2,3,4], m = 2, n = 2 Output: [[1,2],[3,4]] Explanation: The constructed 2D array should contain 2 rows and 2 columns. The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array. The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.
Example 2:
Input: original = [1,2,3], m = 1, n = 3 Output: [[1,2,3]] Explanation: The constructed 2D array should contain 1 row and 3 columns. Put all three elements in original into the first row of the constructed 2D array.
Example 3:
Input: original = [1,2], m = 1, n = 1 Output: [] Explanation: There are 2 elements in original. It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.
Constraints:
1 <= original.length <= 5 104 1 <= original[i] <= 105 1 <= m, n <= 4 104