Spiral Matrix

Leetcode

Problem Statement

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

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();

        vector<int> arr;

        int left = 0, right = m - 1, top = 0, bottom = n - 1; // bounds - inclusive
        int dir = 0; // 0 -> right, 1 -> down, 2 -> left, 3 -> up
        int c = 0;
        int row = 0, col = 0;
        while (c < n * m) { // left < right && top < bottom
            if (dir == 0) {
                for (int i = left; i <= right; i++) {
                    arr.push_back(matrix[row][i]);
                    c++;
                }
                top++;
                col = right;
            } else if (dir == 1) {
                for (int i = top; i <= bottom; i++) {
                    arr.push_back(matrix[i][col]);
                    c++;
                }
                right--;
                row = bottom;
            } else if (dir == 2) {
                for (int i = right; i >= left; i--) {
                    arr.push_back(matrix[row][i]);
                    c++;
                }
                bottom--;
                col = left;
            } else {
                for (int i = bottom; i >= top; i--) {
                    arr.push_back(matrix[i][col]);
                    c++;
                }
                row = top;
                left++;
            }

            dir++;
            dir = dir % 4;
        }

        return arr;
    }
};