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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0054_spiral_matrix;

// #Medium #Top_Interview_Questions #Array #Matrix #Simulation #Programming_Skills_II_Day_8
// #Level_2_Day_1_Implementation/Simulation #Udemy_2D_Arrays/Matrix
// #2023_08_11_Time_0_ms_(100.00%)_Space_41_MB_(9.67%)

import java.util.ArrayList;
import java.util.List;

/**
 * 54 - Spiral Matrix\.
 *
 * Medium
 *
 * Given an `m x n` `matrix`, return _all elements of the_ `matrix` _in spiral order_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg)
 *
 * **Input:** matrix = \[\[1,2,3],[4,5,6],[7,8,9]]
 *
 * **Output:** [1,2,3,6,9,8,7,4,5] 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg)
 *
 * **Input:** matrix = \[\[1,2,3,4],[5,6,7,8],[9,10,11,12]]
 *
 * **Output:** [1,2,3,4,8,12,11,10,9,5,6,7] 
 *
 * **Constraints:**
 *
 * *   `m == matrix.length`
 * *   `n == matrix[i].length`
 * *   `1 <= m, n <= 10`
 * *   `-100 <= matrix[i][j] <= 100`
**/
public class Solution {
    public List spiralOrder(int[][] matrix) {
        List list = new ArrayList<>();
        int r = 0;
        int c = 0;
        int bigR = matrix.length - 1;
        int bigC = matrix[0].length - 1;
        while (r <= bigR && c <= bigC) {
            for (int i = c; i <= bigC; i++) {
                list.add(matrix[r][i]);
            }
            r++;
            for (int i = r; i <= bigR; i++) {
                list.add(matrix[i][bigC]);
            }
            bigC--;
            for (int i = bigC; i >= c && r <= bigR; i--) {
                list.add(matrix[bigR][i]);
            }
            bigR--;
            for (int i = bigR; i >= r && c <= bigC; i--) {
                list.add(matrix[i][c]);
            }
            c++;
        }
        return list;
    }
}