g0401_0500.s0498_diagonal_traverse.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g0401_0500.s0498_diagonal_traverse;

// #Medium #Array #Matrix #Simulation #2022_07_24_Time_4_ms_(59.46%)_Space_55.6_MB_(10.90%)

/**
 * 498 - Diagonal Traverse\.
 *
 * Medium
 *
 * Given an `m x n` matrix `mat`, return _an array of all the elements of the array in a diagonal order_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/04/10/diag1-grid.jpg)
 *
 * **Input:** mat = \[\[1,2,3],[4,5,6],[7,8,9]]
 *
 * **Output:** [1,2,4,7,5,3,6,8,9]
 *
 * **Example 2:**
 *
 * **Input:** mat = \[\[1,2],[3,4]]
 *
 * **Output:** [1,2,3,4]
 *
 * **Constraints:**
 *
 * *   `m == mat.length`
 * *   `n == mat[i].length`
 * *   1 <= m, n <= 10⁴
 * *   1 <= m * n <= 10⁴
 * *   -10⁵ <= mat[i][j] <= 10⁵
**/
public class Solution {
    public int[] findDiagonalOrder(int[][] mat) {
        int m = mat.length;
        int n = mat[0].length;
        int[] output = new int[m * n];
        int idx = 0;
        for (int diag = 0; diag <= m + n - 2; ++diag) {
            if (diag % 2 == 0) {
                for (int k = Math.max(0, diag - m + 1); k <= Math.min(diag, n - 1); ++k) {
                    output[idx++] = mat[diag - k][k];
                }
            } else {
                for (int k = Math.max(0, diag - n + 1); k <= Math.min(diag, m - 1); ++k) {
                    output[idx++] = mat[k][diag - k];
                }
            }
        }
        return output;
    }
}