题目

题解

对于 3 * 3 的矩阵，

m[0][0] = 1, num = 2;
m[0][1] = 2, num = 3;
m[0][2] = 3, num = 4;
top = 1;

m[1][2] = 4, num = 5;
m[2][2] = 5, num = 6;
right = 1;

m[2][1] = 6, num = 7;
m[2][0] = 7, num = 8;
bottom = 1;

m[1][0] = 8, num = 9;
left = 1;

m[1][1] = 9, num = 10;
top = 2;

right = 0;

bottom = 0;

left = 2;

结束.

代码

import java.util.*;

public class code59 {

    public static int[][] generateMatrix(int n) {
        int left = 0, top = 0, right = n - 1, bottom = n - 1;
        int m[][] = new int[n][n];
        int tar = n * n;
        int num = 1;
        while (num <= tar) {
            // left to right.
            for (int i = left; i <= right; i++) {
                m[top][i] = num;
                num++;
            }
            top++;
            // top to bottom.
            for (int i = top; i <= bottom; i++) {
                m[i][right] = num;
                num++;
            }
            right--;
            // right to left.
            for (int i = right; i >= left; i--) {
                m[bottom][i] = num;
                num++;
            }
            bottom--;
            // bottom to top.
            for (int i = bottom; i >= top; i--) {
                m[i][left] = num;
                num++;
            }
            left++;
        }
        return m;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNextInt()) {
            int n = sc.nextInt();
            int res[][] = generateMatrix(n);
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    System.out.printf("%2d ", res[i][j]);
                }
                System.out.println();
            }
        }
    }

}

参考

Spiral Matrix II （模拟法，设定边界，代码简短清晰）