Baozi Leetcode solution 59: Sprial Matrix II

 

Problem Statement 

 

Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

Example:

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

 Problem link

Video Tutorial

You can find the detailed video tutorial here

• Youtube
• B站

 

Thought Process

Straight forward question, could use Spiral Matrix exact same recursion algorithm to solve this, you can also watch that video tutorial here

 

Still needs pay attribute to this guy, had another great solution:

https://leetcode.com/problems/spiral-matrix-ii/discuss/22282/4-9-lines-Python-solutions

Solutions

 

public int[][] generateMatrix(int n) {
    assert n >= 0;

    int[][] res = new int[n][n];

    this.generateMatrixHelper(0, 0, n, n, res, 1);
    return res;
}

private void generateMatrixHelper(
        int i,
        int j,
        int rowSize,
        int colSize,
        int[][] matrix,
        int num
) {
    if (rowSize <= 0 || colSize <= 0) {
        return;
    }

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

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

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

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

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

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

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

 

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

• Spiral Matrix