10K's

Question

Given an m x n matrix, return all elements of the matrix in spiral order.

Example:

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

Algorithm

No trick question, implement step by step according to intuition.

Code

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