LeetCode //C - 59. Spiral Matrix II

59. Spiral Matrix II

Given a positive integer n, generate an n x n matrix filled with elements from 1 to $n^2$ in spiral order.
 

Example 1:

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

Example 2:

Input: n = 1
Output: [[1]]

Constraints:

• 1 <= n <= 20

From: LeetCode
Link: 59. Spiral Matrix II

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Solution:

Ideas：

1. Initialize Variables:

• Initialize the 2D array (the matrix) to be returned.
• Initialize pointers for the number of rows and the size of each row to be returned.
• Initialize variables for the top, bottom, left, and right boundaries of the current “layer” of the spiral.
• Initialize a variable val to 1 for the next value to be inserted into the matrix.

2. Loop Until Done:

• Loop as long as top <= bottom and left <= right.

3. Fill Values:

• Top Row: Fill the top row from left to right and then increment the top boundary.
• Right Column: Fill the right column from top to bottom and then decrement the right boundary.
• Bottom Row: Fill the bottom row from right to left (if top <= bottom) and then decrement the bottom boundary.
• Left Column: Fill the left column from bottom to top (if left <= right) and then increment the left boundary.

4. Increment Value: After filling each cell, increment the value of val for the next cell.

5. Return the Filled Matrix: Return the 2D array once it’s completely filled.

Code:

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** generateMatrix(int n, int* returnSize, int** returnColumnSizes){
    
    
    // Initialize the output array and sizes
    int** output = (int**)malloc(n * sizeof(int*));
    *returnColumnSizes = (int*)malloc(n * sizeof(int));
    for (int i = 0; i < n; ++i) {
    
    
        output[i] = (int*)malloc(n * sizeof(int));
        (*returnColumnSizes)[i] = n;
    }
    *returnSize = n;

    // Initialize variables for boundaries and direction
    int top = 0, bottom = n - 1, left = 0, right = n - 1;
    int val = 1;

    // Loop through the matrix to fill values
    while (top <= bottom && left <= right) {
    
    
        // Fill top row
        for (int i = left; i <= right; ++i) {
    
    
            output[top][i] = val++;
        }
        ++top;

        // Fill right column
        for (int i = top; i <= bottom; ++i) {
    
    
            output[i][right] = val++;
        }
        --right;

        // Fill bottom row
        if (top <= bottom) {
    
    
            for (int i = right; i >= left; --i) {
    
    
                output[bottom][i] = val++;
            }
            --bottom;
        }

        // Fill left column
        if (left <= right) {
    
    
            for (int i = bottom; i >= top; --i) {
    
    
                output[i][left] = val++;
            }
            ++left;
        }
    }

    return output;
}