Question:

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

Example:

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

Thinking:

  • Method:
    • 参考54. Spiral Matrix
    • 结束条件可以增加一项，通过判断当前填入数字的大小判断退出循环。
class Solution {
    public int[][] generateMatrix(int n) {
        if(n == 0) return null;
        int[][] result = new int[n][n];
        int dir = 0; //0: right, 1:down, 2:left, 3:up
        int left = 0; int up = 0; int right = n - 1; int down = n - 1;
        int x = 0; int y = 0;
        int count = 1;
        result[x][y] = count;
        while(up <= down && right >= left && count <= n * n){
            switch(dir){
                case 0:
                    while(++y <= right){
                        result[x][y] = ++count;
                    }
                    up++;
                    y--;
                    dir = 1;
                    break;
                case 1:
                    while(++x <= down){
                        result[x][y] = ++count;
                    }
                    right--;
                    x--;
                    dir = 2;
                    break;
                case 2:
                    while(--y >= left){
                        result[x][y] = ++count;
                    }
                    down--;
                    y++;
                    dir = 3;
                    break;
                case 3:
                    while(--x >= up){
                        result[x][y] = ++count;
                    }
                    left++;
                    x++;
                    dir = 0;
                    break;
                default:
                    break;
            }
        }
        return result;
    }
}

二刷

class Solution {
    public int[][] generateMatrix(int n) {
        if(n == 0) return null;
        int[][] result = new int[n][n];
        int up = 0, down = n - 1, left = 0, right = n - 1;
        int count = 1;
        result[0][0] = 1;
        int dir = 0, x = 0, y = 0;
        while(up <= down && left <= right){
            switch(dir){
                case 0:
                    while(++y <= right)
                        result[x][y] = ++count;
                    y--; up++;
                    dir = 1;
                    break;
                case 1:
                    while(++x <= down)
                        result[x][y] = ++count;
                    x--; right--;
                    dir = 2;
                    break;
                case 2:
                    while(--y >= left)
                        result[x][y] = ++count;
                    ++y; down--;
                    dir = 3;
                    break;
                case 3:
                    while(--x >= up)
                        result[x][y] = ++count;
                    ++x; left++;
                    dir = 0;
                    break;
                default:
                    break;
            }
        }
        return result;
    }
}