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

For example,
Given n = 3,

You should return the following matrix:

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

我在[LeetCode]–54. Spiral Matrix这篇博客中写了visitCircle的方法，其实跟这个大相径庭的，这个里面我写了fillCircle的方法。思想是一样的，也是一圈一圈的往里填。

public class Solution {
    private static int index = 1;

    public int[][] generateMatrix(int n) {
        int[][] matrix = new int[n][n];
        for (int i = 0; i < n; i++)
            if (matrix[i][i] == 0)
                fillCircle(matrix, i, n);
        return matrix;
    }

    public void fillCircle(int[][] a, int m, int n) {
        for (int i = m; i < n - m; i++) {
            a[m][i] = index++;
        }
        for (int i = 1 + m; i < n - m; i++) {
            a[i][n - m - 1] = index++;
        }

        if (n - 2 * m == 1 || n - 2 * m == 1)
            return;

        for (int i = n - 2 - m; i >= m; i--) {
            a[n - m - 1][i] = index++;
        }
        for (int i = n - m - 2; i > m; i--) {
            a[i][m] = index++;
        }
    }
}

但是这个没有AC理由是输入2的时候答案是不正确的。但是我runCode的时候答案又是正确的，而且我在我eclipse里面完全正确啊。感觉是个bug吧。

再来个可以AC的算法。

public int[][] generateMatrix1(int n) {
        if (n < 0) {
            return null;
        }

        int[][] result = new int[n][n];

        int xStart = 0;
        int yStart = 0;
        int num = 1;

        while (n > 0) {
            if (n == 1) {
                result[yStart][xStart] = num++;
                break;
            }

            for (int i = 0; i < n - 1; i++) {
                result[yStart][xStart + i] = num++;
            }

            for (int i = 0; i < n - 1; i++) {
                result[yStart + i][xStart + n - 1] = num++;
            }

            for (int i = 0; i < n - 1; i++) {
                result[yStart + n - 1][xStart + n - 1 - i] = num++;
            }

            for (int i = 0; i < n - 1; i++) {
                result[yStart + n - 1 - i][xStart] = num++;
            }

            xStart++;
            yStart++;
            n = n - 2;
        }

        return result;
    }