Leetcode. Spiral Matrix

Given a matrix of m x n elements (m rows, n columns), return all elements of the Matri X in Spiral Order.

For example,
Given the following matrix:

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

You should return [1,2,3,6,9,8,7,4,5] .

Use left, top, right, bottom to record boundary conditions.

1vector<int> Spiralorder (vector<vector<int> > &matrix)2     {3vector<int>ret;4         intm =matrix.size ();5         if(M <=0)6             returnret;7         intn = matrix[0].size (), i =0, j =0;8         intleft =0, right = N-1, top =0, bottom = M-1;9         Ten          while(true) One         { Ai = top, j =Left ; -              while(J <=Right ) -Ret.push_back (matrix[i][j++]); the             if(++top > Bottom) Break; -              -i = top, j =Right ; -              while(I <=bottom) +Ret.push_back (matrix[i++][j]); -             if(--right < left) Break; +              Ai = bottom, j =Right ; at              while(J >=Left ) -Ret.push_back (matrix[i][j--]); -             if(--bottom < top) Break; -              -i = bottom, j =Left ; -              while(I >=top) inRet.push_back (matrix[i--][j]); -             if(++left > right) Break; to         } +          -         returnret; the}

Given an integer n, generate a square matrix filled with elements from 1 to n2 in Spiral order.

For example,
Given N = 3 ,

You should return the following matrix:

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

1vector<vector<int> > Generatematrix (intN)2     {3vector<vector<int> >Matrix;4         if(N <=0)5             returnMatrix;6         intleft =0, right = N-1, top =0, bottom = n-1, I, j, k =1;7vector<int>Line (n);8          for(i =0; I < n; i++)9 Matrix.push_back (line);Ten          while(true) One         { Ai = top, j =Left ; -              while(J <=Right ) -Matrix[i][j++] = k++; the             if(++top > Bottom) Break; -              -i = top, j =Right ; -              while(I <=bottom) +MATRIX[I++][J] = k++; -             if(--right < left) Break; +              Ai = bottom, j =Right ; at              while(J >=Left ) -matrix[i][j--] = k++; -             if(--bottom < top) Break; -              -i = bottom, j =Left ; -              while(I >=top) inMATRIX[I--][J] = k++; -             if(++left > right) Break; to         } +          -         returnMatrix; the}

Leetcode. Spiral Matrix