Spiral Matrix (#54 Leetcode)

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

 

Program 

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int left = 0, right = matrix[0].size(); // Columns
        int top = 0, bottom = matrix.size(); // Rows
        vector<int> res; 
             
        while (left < right && top < bottom) {
            
            for (int i = left; i < right; i++) {
                // Top row
                res.push_back(matrix[top][i]);
            }
            top++;
            
            for (int i = top; i < bottom; i++) {
                // Right Column
                res.push_back(matrix[i][right - 1]);
            }
            right--;
            
            if (left >= right || top >= bottom) {
                break;
            }
        
            for (int i = right-1; i >= left; i--) {
                // Bottom Row   
                res.push_back(matrix[bottom-1][i]);
                
            }
            bottom--;
            
            for (int i = bottom-1; i >= top; i--) {
                // Left Column
                res.push_back(matrix[i][left]);
            }
            left++;
        
        }
        
        return res;
    }
};