59. Spiral matrix (array)

  subject ：

Given a positive integer  n, Generate a include 1 To  n^2  All the elements , A square matrix in which the elements are spirally arranged in a clockwise order .

Example :

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

public int[][] generateMatrix(int n){
        // Create a 2D array 
        int [][] res = new int[n][n];
        // Define the starting position of each cycle 
        int startX =0,startY=0;
        // How many times per cycle , for example n It's odd 3, that loop=3/2=1, Just a cycle , The in the middle of the matrix only needs to be handled separately 
        int loop = n/2;
        // In the middle of the matrix , for example ：n by 3, The middle position is [1,1];
        int mid = n/2;
        // Used to assign a value to each space in the matrix 
        int count =1;
        // Every cycle , You need to control the length of each edge traversal 
        int offset =1;
        int i,j;
        while(loop >0){
            i = startX;
            j = startY;

            // The following four for Is to simulate a turn 
            // Analog fill uplink from left to right （ Left closed right away ）
            for (j = startY; j <startY+n-offset ; j++) {
                res[startX][j] = count++;
            }
            // The simulation fills the right column from top to bottom （ Left closed right away ）
            for (i = startX; i <startX+n-offset ; i++) {
                res[i][j] = count++;
            }
            // The simulation fills down from right to left （ Left closed right away ）
            for (; j>startY;j--){
                res[i][j] = count++;
            }
            // Simulate filling the left column from bottom to top （ Left closed right away ）
            for (; i>startX;i--){
                res[i][j] = count++;
            }

            // At the beginning of the second lap , The starting position shall be added with 1,
            startX++;
            startY++;

            //offset Control the length of each edge traversal in each circle 
            offset+=2;
            loop--;
        }
        // If n It's odd , You need to assign a separate value to the middle of the matrix 
        if(n%2==1){
            res[mid][mid] = count;
        }
        return res;
    }

public static void main(String[] args) {
        Luoxuanjuzhen luoxuanjuzhen = new Luoxuanjuzhen();
        int [][]res = luoxuanjuzhen.generateMatrix(4);
        for (int [] x:res) {
            for (int y:x) {
                System.out.print(y+" ");
            }
        }
    }

Output results ：

1 2 3 4 12 13 14 5 11 16 15 6 10 9 8 7 

Time complexity ：O（n）