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 n^2 n2 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

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:

c 复制代码

/**
 * 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;
}