Matrices

Spiral Matrix

medium

DESCRIPTION (inspired by Leetcode.com)

Write a function to traverse an m x n matrix in spiral order and return all elements in a single list. The traversal should start from the top left corner and proceed clockwise, spiraling inward until every element has been visited.

Input:

matrix = [
    [0,1,2],
    [3,4,5],
    [6,7,8]
]

Output:

[0,1,2,5,8,7,6,3,4]

Explanation: The elements of the matrix are returned in the order they are visited in a clockwise spiral starting from the top left corner.

Code Editor

Java

public class Solution {
    public List&amp;lt;Integer&amp;gt; spiral_order(int[][] matrix) {
        // Your code goes here
    }
}

Run your code to see results here

Have suggestions or found something wrong?

Explanation

This solution uses 4 steps to traverse the matrix in spiral order:

1. Top Row

The first step is to pop the first row of the matrix, while copying those elements into the result array from left to right.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

initialize result

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2. Right Column

Next, we traverse the right column of the matrix, popping the last element of each remaining row in matrix and copying those elements into the result array from top to bottom.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

traverse top row

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3. Bottom Row

Next, we traverse the bottom row of the matrix, popping the last row of matrix and copying those elements into the result array from right to left.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

traverse right column

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4. Left Column

Finally, we traverse the left column of the matrix, popping the first element of each remaining row in matrix and copying those elements into the result array from bottom to top.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

traverse bottom row

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Repeat

At this point, we have traversed the perimeter of the original matrix in a clockwise spiral order. We repeat the process for the inner submatrix, until there are no more elements left to traverse.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

traverse left column

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Solution

Visualization

Java

Try these examples:


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

spiral matrix

0 / 8

What is the time complexity of this solution?

O(m * n)

O(N + Q)

O(n log n)

O(log n)

Checking progress…

Matrices

Spiral Matrix

medium

DESCRIPTION (inspired by Leetcode.com)

Input:

matrix = [
    [0,1,2],
    [3,4,5],
    [6,7,8]
]

Output:

[0,1,2,5,8,7,6,3,4]

Explanation: The elements of the matrix are returned in the order they are visited in a clockwise spiral starting from the top left corner.

Code Editor

Java

public class Solution {
    public List&amp;lt;Integer&amp;gt; spiral_order(int[][] matrix) {
        // Your code goes here
    }
}

Run your code to see results here

Have suggestions or found something wrong?

Explanation

This solution uses 4 steps to traverse the matrix in spiral order:

1. Top Row

The first step is to pop the first row of the matrix, while copying those elements into the result array from left to right.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

initialize result

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2. Right Column

Next, we traverse the right column of the matrix, popping the last element of each remaining row in matrix and copying those elements into the result array from top to bottom.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

traverse top row

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3. Bottom Row

Next, we traverse the bottom row of the matrix, popping the last row of matrix and copying those elements into the result array from right to left.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

traverse right column

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4. Left Column

Finally, we traverse the left column of the matrix, popping the first element of each remaining row in matrix and copying those elements into the result array from bottom to top.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

traverse bottom row

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Repeat

At this point, we have traversed the perimeter of the original matrix in a clockwise spiral order. We repeat the process for the inner submatrix, until there are no more elements left to traverse.

Visualization

Java


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

traverse left column

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Solution

Visualization

Java

Try these examples:


public List<Integer> spiralOrder(List<List<Integer>> matrix) {
    List<Integer> result = new ArrayList<>();
    while (!matrix.isEmpty()) {
        result.addAll(matrix.remove(0));
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (List<Integer> row : matrix) {
                result.add(row.remove(row.size() - 1));
            }
        }
        if (!matrix.isEmpty()) {
            List<Integer> lastRow = matrix.remove(matrix.size() - 1);
            Collections.reverse(lastRow);
            result.addAll(lastRow);
        }
        if (!matrix.isEmpty() && !matrix.get(0).isEmpty()) {
            for (int i = matrix.size() - 1; i >= 0; i--) {
                result.add(matrix.get(i).remove(0));
            }
        }
    }
    return result;
}

spiral matrix

0 / 8

What is the time complexity of this solution?

O(m * n)

O(N + Q)

O(n log n)

O(log n)

Checking progress…

Top pages

Top content

Spiral Matrix

DESCRIPTION (inspired by Leetcode.com)

Explanation

1. Top Row

2. Right Column

3. Bottom Row

4. Left Column

Repeat

Solution

What is the time complexity of this solution?

Top pages

Top content

Spiral Matrix

DESCRIPTION (inspired by Leetcode.com)

Explanation

1. Top Row

2. Right Column

3. Bottom Row

4. Left Column

Repeat

Solution

What is the time complexity of this solution?