59. Spiral Matrix II

Explanation

To generate a spiral matrix, we can simulate the process of "walking" along the borders of the matrix and filling in the elements. We keep track of the current position and direction of movement. As we move along the borders, we update the matrix with the corresponding values.

Algorithm

Initialize the matrix with all zeros and set the current value to 1.
Define four boundaries: top, bottom, left, and right, which represent the boundaries of the current spiral loop.
Iterate over each value from 1 to n^2, filling the matrix in the spiral order.
Update the current position and direction based on the boundaries.
Return the generated matrix.

Time Complexity

The time complexity of this algorithm is O(n^2) as we visit each cell in the n x n matrix exactly once.

Space Complexity

The space complexity of this algorithm is O(n^2) as we are creating an n x n matrix to store the values.

View Leetcode Problem

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] matrix = new int[n][n];
        int top = 0, bottom = n - 1, left = 0, right = n - 1;
        int value = 1;

        while (value <= n * n) {
            for (int i = left; i <= right && value <= n * n; i++) {
                matrix[top][i] = value++;
            }
            top++;

            for (int i = top; i <= bottom && value <= n * n; i++) {
                matrix[i][right] = value++;
            }
            right--;

            for (int i = right; i >= left && value <= n * n; i--) {
                matrix[bottom][i] = value++;
            }
            bottom--;

            for (int i = bottom; i >= top && value <= n * n; i--) {
                matrix[i][left] = value++;
            }
            left++;
        }

        return matrix;
    }
}

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