54. Spiral Matrix - DevCodeEx

Explanation:

To traverse the matrix in a spiral order, we can repeatedly extract the outermost ring of the matrix until the matrix is empty. We keep track of the boundaries of the current ring and move in four directions (right, down, left, up) while adding elements to the result list.

Algorithmic Idea:
1. Initialize four variables to keep track of the boundaries of the current ring: top, bottom, left, right.
2. Traverse the matrix in a spiral order by moving right, down, left, and up, while updating the boundaries and adding elements to the result list.
3. Repeat the process until the matrix is empty.
Time Complexity: O(m * n) where m is the number of rows and n is the number of columns in the matrix.
Space Complexity: O(1) excluding the result list.

:

View Leetcode Problem

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> result = new ArrayList<>();
        if (matrix == null || matrix.length == 0) {
            return result;
        }
        
        int top = 0, bottom = matrix.length - 1;
        int left = 0, right = matrix[0].length - 1;
        
        while (top <= bottom && left <= right) {
            for (int i = left; i <= right; i++) {
                result.add(matrix[top][i]);
            }
            top++;
            
            for (int i = top; i <= bottom; i++) {
                result.add(matrix[i][right]);
            }
            right--;
            
            if (top <= bottom) {
                for (int i = right; i >= left; i--) {
                    result.add(matrix[bottom][i]);
                }
                bottom--;
            }
            
            if (left <= right) {
                for (int i = bottom; i >= top; i--) {
                    result.add(matrix[i][left]);
                }
                left++;
            }
        }
        
        return result;
    }
}

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