Algorithm to Generate the Spiral Matrix in Clock-wise Order

  • 时间：2020-09-12 10:06:27
  • 分类：网络文摘
  • 阅读：137 次

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
4
5

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

Walk and Turn Algorithm to Fill the Matrix in Sprial Clock-wise Order

We start at the top-left corner where we fill number 1, then the initial direction is RIGHT, then we keep walking until we hit the border or the cell has been filled already. Then we turn right, repeatedly doing this until we have finished the matrix.

The special case is the 1×1 matrix, we can just immediately return [1] without walking. The following is the Java implementation of the Clock-wise spiral matrix. In Java, we use Arrays.fill to initialize a one-dimension array. We can use a for loop to initialize a two dimensional array using Arrays.fill.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] res = new int[n][n];
        for (int[] x: res) {
            Arrays.fill(x, -1);
        }
        if (n <= 1) {
            res[0][0] = 1;
            return res;
        }
        int d[][] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        int x = 0, r = 0, c = 0, num = 1;
        int total = n * n;
        res[0][0] = 1;
        while (num < total) {
            int nr = r + d[x][0];
            int nc = c + d[x][1];
            if (nr >= n || nc >= n || nr < 0 || nc < 0 || res[nr][nc] != -1) {
                x = (x + 1) % 4;
            } else {
                r = nr;
                c = nc;
                res[r][c] = ++ num;
            }
        }
        return res;        
    }
}
class Solution {
    public int[][] generateMatrix(int n) {
        int[][] res = new int[n][n];
        for (int[] x: res) {
            Arrays.fill(x, -1);
        }
        if (n <= 1) {
            res[0][0] = 1;
            return res;
        }
        int d[][] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        int x = 0, r = 0, c = 0, num = 1;
        int total = n * n;
        res[0][0] = 1;
        while (num < total) {
            int nr = r + d[x][0];
            int nc = c + d[x][1];
            if (nr >= n || nc >= n || nr < 0 || nc < 0 || res[nr][nc] != -1) {
                x = (x + 1) % 4;
            } else {
                r = nr;
                c = nc;
                res[r][c] = ++ num;
            }
        }
        return res;        
    }
}

Similarly, the following is the C++ version of the Spiral square matrix.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

class Solution {
public:
    vector<vector>int>> generateMatrix(int n) {
        vector<vector<int>> res(n, vector<int>(n, -1));
        if (n <= 1) {
            return vector<vector<int>>(1, {1});
        }
        int d[][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        int x = 0, r = 0, c = 0, num = 1;
        int total = n * n;
        res[0][0] = 1;
        while (num < total) {
            int nr = r + d[x][0];
            int nc = c + d[x][1];
            if (nr >= n || nc >= n || nr < 0 || nc < 0 || res[nr][nc] != -1) {
                x = (x + 1) % 4;
            } else {
                r = nr;
                c = nc;
                res[r][c] = ++ num;
            }
        }
        return res;
    }
};
class Solution {
public:
    vector<vector>int>> generateMatrix(int n) {
        vector<vector<int>> res(n, vector<int>(n, -1));
        if (n <= 1) {
            return vector<vector<int>>(1, {1});
        }
        int d[][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        int x = 0, r = 0, c = 0, num = 1;
        int total = n * n;
        res[0][0] = 1;
        while (num < total) {
            int nr = r + d[x][0];
            int nc = c + d[x][1];
            if (nr >= n || nc >= n || nr < 0 || nc < 0 || res[nr][nc] != -1) {
                x = (x + 1) % 4;
            } else {
                r = nr;
                c = nc;
                res[r][c] = ++ num;
            }
        }
        return res;
    }
};

Syntaxally much the same, both implementations run O(N^2) time and space complexity.

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
王叔叔的手表　 韦达未卜先知　 什么是对数——对数的发展简史　 棋子数目问题　 丁谓施工——统筹学的古代应用　 最终得到的一位数　 学会换个思路　 一道有关分数的问题　 求两地距离　 一个追及问题　
评论列表
添加评论