[Leetcode 98] 54 Spiral Matrix

Problem:

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]

You should return [1,2,3,6,9,8,7,4,5].

 

Analysis:

The problem is a simulation problem. With the help of a direction variable, we can go through the matrix. When a new position is not valid, then we must change the direction.

For a index (i, j) to be valid, the following creteria must be satisfied:

1. i >=0 && i < row

2. j>=0 && j < col

3. m[i][j] is not visited

 

Direction changes from right -> down -> left -> up.

 

Code:

class Solution {
public:
    int row, col, MIN = (1<<31);

    vector<int> spiralOrder(vector<vector<int> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> res;
        
        if (matrix.size() == 0)
            return res;
            
        row = matrix.size(), col = matrix[0].size();
        int dir = 0, size = row * col;
        int row_dir[4] = {0, 1, 0, -1};
        int col_dir[4] = {1, 0, -1, 0};
        
        int i=0, j=-1, total = 0;
        while (true) {
            i += row_dir[dir];
            j += col_dir[dir];
            
            if (isValid(i, j, matrix)) {
                res.push_back(matrix[i][j]);
                matrix[i][j] = MIN;
                total += 1;
                if (total == size) break;
            } else {
                // reset to last valid position
                i -= row_dir[dir];
                j -= col_dir[dir];
                dir = (dir + 1) % 4;
            }
        }
        
        return res;
    }
    
    bool isValid(int i, int j, vector<vector<int> > & m) {
        return (i < row && i>=0) &&( j>=0 && j < col) && (m[i][j] != MIN);
    }
};