spiral-matrix-ii

　　

/**
 * Given an integer n, generate a square matrix filled with elements from 1 to n 2 in spiral order.
 * For example,
 * Given n =3,
 * You should return the following matrix:
 * [
 *  [ 1, 2, 3 ],
 *  [ 8, 9, 4 ],
 *  [ 7, 6, 5 ]
 * ]
 *
 * 给定一个整数n，生成一个以螺旋顺序填充从1到n 2元素的正方形矩阵。
 * 例如，
 * 如果n=3，
 * 您应该返回以下矩阵：
 * You should return the following matrix:
 * [
 *  [ 1, 2, 3 ],
 *  [ 8, 9, 4 ],
 *  [ 7, 6, 5 ]
 * ]
 */

/**
 * Given an integer n, generate a square matrix filled with elements from 1 to n 2 in spiral order.
 * For example,
 * Given n =3,
 * You should return the following matrix:
 * [
 *  [ 1, 2, 3 ],
 *  [ 8, 9, 4 ],
 *  [ 7, 6, 5 ]
 * ]
 *
 * 给定一个整数n，生成一个以螺旋顺序填充从1到n 2元素的正方形矩阵。
 * 例如，
 * 如果n=3，
 * 您应该返回以下矩阵：
 * You should return the following matrix:
 * [
 *  [ 1, 2, 3 ],
 *  [ 8, 9, 4 ],
 *  [ 7, 6, 5 ]
 * ]
 */

public class Main46 {
    public static void main(String[] args) {
        int n = 5;
        int[][] matrix = generateMatrix(n);
        for (int i=0;i<n;i++) {
            for (int j=0;j<n;j++) {
                System.out.print(matrix[i][j]+"    ");
            }
            System.out.println("");
        }
    }

    public static int[][] generateMatrix(int n) {
        int[][] matrix = new int[n][n];
        int count = 1;
        int start = 0;
        while (count <= n*n) {
            int endX = n-1-start;   //列
            int endY = n-1-start;   //行

            for (int i=start;i<=endX;i++) {
                matrix[start][i] = count;
                count++;
            }
            for (int i=start+1;i<=endY;i++) {
                matrix[i][endY] = count;
                count++;
            }
            for (int i=endX-1;i>=start && endY>start;i--) {
                matrix[endY][i] = count++;
            }
            for (int i=endY-1;i>=start+1 && endX>start;i--) {
                matrix[i][start] = count++;
            }
            start++;
        }
        return matrix;
    }

}