lc-54

//Given an m x n matrix, return all elements of the matrix in spiral order.
//
//
// Example 1:
//
//
//Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
//Output: [1,2,3,6,9,8,7,4,5]
//
//
// Example 2:
//
//
//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
//
// Related Topics Array Matrix Simulation
// 👍 7991 👎 873


//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        int[][] dirs = {{0,1}, {1,0}, {0, -1}, {-1, 0}};
        if (matrix == null || matrix[0] == null || matrix.length == 0 || matrix[0].length == 0) {
            return null;
        }

        boolean[][] visited = new boolean[matrix.length][matrix[0].length];
        int curDirIndex = 0;

        int i = 0, j = 0;
        List<Integer> result = new ArrayList<>();

        while (i >= 0 && i < matrix.length && j >= 0 && j < matrix[0].length && !visited[i][j]) {
            result.add(matrix[i][j]);
            visited[i][j] = true;
            curDirIndex = getNextDirIndex(curDirIndex, i, j, matrix, visited);
            i = i + dirs[curDirIndex][0];
            j = j + dirs[curDirIndex][1];
        }
        return result;
    }

    private int getNextDirIndex(int curDirIndex, int i, int j, int[][] matrix, boolean[][] visited) {
        if (curDirIndex == 0 && (j == matrix[0].length - 1 || visited[i][j + 1])) {
            return 1;
        } else if (curDirIndex == 1 && (i == matrix.length - 1 || visited[i + 1][j])) {
            return 2;
        } else if (curDirIndex == 2 && (j == 0 || visited[i][j - 1])) {
            return 3;
        } else if (curDirIndex == 3 && (i == 0 || visited[i - 1][j])) {
            return 0;
        } else {
            return curDirIndex;
        }
    }
}
//leetcode submit region end(Prohibit modification and deletion)