Baozi Leetcode solution 54: Sprial Matrix

Problem Statement 

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

Example 1:

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

Example 2:

Input:
[
  [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]

 Problem link

Video Tutorial

You can find the detailed video tutorial here

• Youtube
• B站

 

Thought Process

It's a straight forward implementation question, we can simply just simulate the spiral order and print. You can choose to do it either iteratively (see reference to download the official Leetcode solution) or use recursion.

 

There is this awesome one line solution from this guy which is pretty insane.

def spiralOrder(self, matrix):
    return matrix and list(matrix.pop(0)) + self.spiralOrder(zip(*matrix)[::-1])

https://leetcode.com/problems/spiral-matrix/discuss/20571/1-liner-in-Python-%2B-Ruby

 

Solutions

public List<Integer> spiralOrder(int[][] matrix) {
    List<Integer> res = new ArrayList<>();

    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return res;
    }

    this.printSpiralOrder(0, 0, matrix.length, matrix[0].length, res, matrix);

    return res;
}

// printSpiralOrder is a poor name, function starts with verb, printSpiralOrder, class is noun, function is verb
private void printSpiralOrder(int i, int j, int rowSize, int colSize, List<Integer> res, int[][] matrix) {
    if (rowSize <= 0 || colSize <= 0) {
        return;
    }

    if (rowSize == 1 && colSize == 1) {
        res.add(matrix[i][j]);
        return;
    }
    if (rowSize == 1) {
        for (int k = j; k < j + colSize; k++) {
            res.add(matrix[i][k]);
        }
        return;
    }

    if (colSize == 1) {
        for (int k = i; k < i + rowSize; k++) {
            res.add(matrix[k][j]);
        }
        return;
    }

    // do the spiral
    for (int k = j; k < j + colSize; k++) {
        res.add(matrix[i][k]);
    }

    for (int k = i + 1; k < i + rowSize; k++) {
        res.add(matrix[k][j + colSize - 1]);
    }

    for (int k = j + colSize - 2; k >= i; k--) {
        res.add(matrix[i + rowSize - 1][k]);
    }

    for (int k = i + rowSize - 2; k > i; k--) {   // both the start and end need to be i, j, and also care about length
        res.add(matrix[k][j]);
    }

    this.printSpiralOrder(i + 1, j + 1, rowSize - 2, colSize - 2, res, matrix);
} 

Simulation using Recursion

Time Complexity: O(M*N) where M, N is row and col of matrix

Space Complexity: O(M*N) since we used list to store the result, where M, N is row and col of matrix

 

 

References

• Leetcode official solution (download pdf)