leetcode 59. Spiral matrix II

leetcode 59. Spiral matrix Ⅱ

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Title Description
Give you a positive integer n , Generate a include 1 To n2 All the elements , And the elements are arranged in a clockwise spiral order n x n square matrix matrix .

Input n=3

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

Answer key

The main thing is to find out the conditions for each turn , Then keep turning （ In whole circles ）

/* * @lc app=leetcode.cn id=59 lang=cpp * * [59]  Spiral matrix  II */

// @lc code=start
class Solution
{
    
public:
    vector<vector<int>> generateMatrix(int n)
    {
    
        vector<vector<int>> res(n, vector<int>(n, 0)); // Define an array for storing data 
        int startx = 0, starty = 0;                    // Definition x and y Axis 
        int loop = n / 2;                              // Calculate the number of times the created matrix needs to be cycled 
        int mid = n / 2;                               // Calculate the element subscript of the intermediate element 
        int count = 1;                                 // Count assignment 
        int offset = 1;                                // Control the length of the edge traversal field of each layer 
        int i, j;
        while (loop--)
        {
    
            // The data of the above row is filled in 
            for (j = starty; j < starty + n - offset; j++)
            {
    
                res[startx][j] = count++;
                /* code */
            }
            // Filling of data on the right 
            for (i = startx; i < startx + n - offset; i++)
            {
    
                res[i][j] = count++;
            }
            // Fill in the following data 
            for (; j > starty; j--)
            {
    
                res[i][j] = count++;
                /* code */
            }
            // Filling of data on the left 
            for (; i > startx; i--)
            {
    
                res[i][j] = count++;
                /* code */
            }
            // Shrink the subscript in one circle for each circle filled 
            starty++;
            startx++;
            // Control the traversal times of each edge, and cannot reach the last block 
            offset += 2;
            /* code */
        }
        if (n % 2)
        {
    
            res[mid][mid] = count;
        }
        return res;
    }
};
// @lc code=end

count;
}
return res;
}
};
// @lc code=end