2D array theme

1. Power button 48 Rotated image

Rotated image ： Rotate the image clockwise 90 degree

step1. First, put this n*n Matrix according to diagonal （ The diagonal from the top left to the bottom right , Instead of the diagonal line from the upper right corner to the lower left corner ） Symmetry . For example, the first row becomes the first column after symmetry

  Diagonal symmetric code ：

       for(int i=0;i<n;i++)
        {
            for(int j=i;j<n;j++)
            {
                int temp=nums[i][j];
                nums[i][j]=nums[j][i];
                nums[j][i]=temp;
            }
        }

The reason why j from i Start , It means the part with blue line （ The diagonal element of each row and the part to the right of the diagonal element ）

step2： Then invert each row of the matrix

For example, the first line is 1,5,7,4, After reversing, it becomes 4,7,5,1

  After these two steps, it is found that the matrix has rotated clockwise 90 The degree of

The code implementation is as follows ：

class Solution 
{
    public void rotate(int[][] nums) 
    {
        // This problem requires rotating in place , You cannot use another new matrix to install the rotated matrix 
        // Find the side length of this square matrix （ Equal length and width ）
        int n=nums.length;
        
        // First step ： First symmetrically along the diagonal 
        for(int i=0;i<n;i++)
        {
            for(int j=i;j<n;j++)
            {
                int temp=nums[i][j];
                nums[i][j]=nums[j][i];
                nums[j][i]=temp;
            }
        }

        // The second step ： Each line is symmetrical 
        // Take this out matric Every line of the matrix 
        for(int i=0;i<n;i++)
        {
            int p=0,q=n-1;
            while(p<q)
            {
              int temp=nums[i][p];
              nums[i][p]=nums[i][q];
              nums[i][q]=temp;
              p++;
              q--;
            }
        }
    }
}

2. Power button 54  Spiral matrix

class Solution 
{
    public List<Integer> spiralOrder(int[][] nums)
    {
        int m=nums[0].length; // Matrix length 
        int n=nums.length;// Matrix width 

        // Define four boundaries 
        // Note that there m,n Don't write the position of in reverse , It's easy to make a mistake                                                                                   
        int left=0; // What column is the left boundary in 
        int right=m-1; // Which column is the right boundary 
        int up=0; // Which line is the upper boundary 
        int down=n-1;// Which line is the lower boundary 

        List<Integer> result=new ArrayList();
        
        // hold m*n All elements are added to result
        while(result.size()<m*n)
        {
            // It is required to rotate clockwise , So add the upper boundary in turn , Right border , Lower boundary , The number of left bounds 

            // Add the number of upper boundaries 
            if(up<=down)
            {
                for(int temp=left;temp<=right;temp++)
                {
                    result.add(nums[up][temp]);
                }
                // Move the upper boundary down 
                up++;
            }

          // Add the number of right boundaries 
            if(left<=right)
            {
                for(int temp=up;temp<=down;temp++)
                {
                    result.add(nums[temp][right]);
                }
                // Move the right border to the left 
                right--;
            }
            
            // Add the number of lower boundaries 
            if(up<=down)
            {
                for(int temp=right;temp>=left;temp--)
                {
                    result.add(nums[down][temp]);
                }
                // Move lower boundary up 
                down--;
            }

            // Add the number of left boundaries 
            if(left<=right)
            {
                for(int temp=down;temp>=up;temp--)
                {
                    result.add(nums[temp][left]);
                }
                // Move left border right 
                left++;
            }
        }
        return result;
    }
}

3.59. Spiral matrix II - Power button （LeetCode）

class Solution 
{
    public int[][] generateMatrix(int n)
    {
        int[][]  result=new int[n][n];

        int up=0;
        int down=n-1;
        int left=0;
        int right=n-1;

        // from 1 Start filling the square matrix , Until n The square of 
        int num=1;

        while(num<=n*n)
        {
            // The square matrix is arranged clockwise 
            // First fill the upper boundary , Then fill the right boundary , Then fill the lower boundary , Finally, fill the left boundary 
            
            // Fill the upper boundary 
            if(up<=down)
            {
                for(int temp=left;temp<=right;temp++)
                {
                  result[up][temp]=num;
                  num++;
                }
                // Move the upper boundary down 
                up++;
            }

            // Fill the right boundary 
            if(left<=right)
            {
                for(int temp=up;temp<=down;temp++)
                {
                  result[temp][right]=num;
                  num++;
                }
                // Move the right border to the left 
                right--;
            }

            // Fill the lower boundary 
            if(up<=down)
            {
                for(int temp=right;temp>=left;temp--)
                {
                  result[down][temp]=num;
                  num++;
                }
                // Move lower boundary up 
                down--;
            }

            // Fill left boundary 
            if(left<=down)
            {
                for(int temp=down;temp>=up;temp--)
                {
                  result[temp][left]=num;
                  num++;
                }
                // Move left border right 
                left++;
            }

        }
        return result;
    }
}