Force buckle 59 Spiral matrix II

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 .

Method ： Set the upper, lower, left and right boundaries  

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        // Use vector Define a two-dimensional array 
        vector<vector<int>> res(n, vector<int>(n, 0)); 

        // Define the rows and columns represented by the top, bottom, left and right , As the controlling factor of the boundary 
        int top = 0;
        int bottom = n - 1;
        int left = 0;
        int right = n - 1;
        int num = 1;  // First element placed 
        int tar = n * n;  // Put the last element 

        // This loop is used to ensure that all elements are put into the matrix 
        // The order in which the elements are placed follows the law of spirals , Clockwise order 
        while(num <= tar) {
            // The upper side of the array is filled from left to right 
            // The line doesn't change , Column change , Which line is it , according to top( Upper side ) Location determines 
            for(int i = left; i <= right; ++i) {
                res[top][i] = num++;     
            }
            ++top;  // Fill up , The upper side of the array is full of one row ,top+1

            // The right side of the array is filled from top to bottom 
            // Row change , The column does not change , Which column is it , according to right( On the right side ) Location determines 
            for(int i = top; i <= bottom; ++i) {
                res[i][right] = num++;
            }
            --right;  // Fill up , The right side of the array is full of a column 

            // The lower side of the array is filled from right to left 
            // The line doesn't change , Column change , Which line is it , according to bottom( Lower side ) Location determines 
            for(int i = right; i >= left; --i) {
                res[bottom][i] = num++;
            }
            --bottom;  // Fill up , The lower side of the array is full of one row 

            // The left side of the array is filled from bottom to top 
            // Row change , The column does not change , Which column is it , according to left( left ) Location determines 
            for(int i = bottom; i >= top; --i) {
                res[i][left] = num++;
            }
            ++left;  // Fill up , The left side of the array is full 
        }
        return res;
    }
    // Time complexity ：O(n^2), among  n  Is a given positive integer . The size of the matrix is  n×n, You need to fill in every element of the matrix .
    // Spatial complexity ：O(1). In addition to the returned matrix , Space complexity is a constant 

};

Answer key ：

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