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：166 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：
八种营养价值很高的“难吃”食物　 健康养生：胡萝卜怎么食用才更营养　 初秋多吃些薯类食物可清肠排毒抗衰　 保健养生：三种薯类食物健康食疗方　 转基因食品推广需尊重消费者知情权　 转基因食品试验不应是一场推销闹剧　 管理总局曝光七个保健食品违法广告　 预防湿疹复发可常吃清热祛湿食物　 调和油乱象：市场价格和行业标准之乱　 消费者该如何识别和选择食用油？　
评论列表
添加评论