*spiral Matrix

title :

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] .

The following:
This problem is a realization problem.

Consider 2 initial conditions, if the matrix has only one row or column, then you do not need to rotate, then output.

All other situations need to be in circles: from left to right, from top to bottom, right to left, bottom to top. Loop from the big circle to the small circle in turn.

The code is as follows:

1  PublicArraylist<integer> Spiralorder (int[] matrix) {2arraylist<integer> result =NewArraylist<integer>();3         if(Matrix = =NULL|| Matrix.length = = 0)4             returnresult;5  6         intm =matrix.length;7         intn = matrix[0].length;8  9         intX=0; Ten         intY=0; One   A          while(M>0 && n>0){ -   -             //if one row/column left, no circle can be formed the             if(m==1){ -                  for(inti=0; i<n; i++){ -Result.add (matrix[x][y++]); -                 } +                  Break; -}Else if(n==1){ +                  for(inti=0; i<m; i++){ AResult.add (matrix[x++][y]); at                 } -                  Break; -             } -   -             //below, process a circle -   in             //Top-move Right -              for(inti=0;i<n-1;i++) toResult.add (matrix[x][y++]); +   -             //Right-move Down the              for(inti=0;i<m-1;i++) *Result.add (matrix[x++][y]); $  Panax Notoginseng             //Bottom-move Left -              for(inti=0;i<n-1;i++) theResult.add (matrix[x][y--]); +   A             //Left-move up the              for(inti=0;i<m-1;i++) +Result.add (matrix[x--][y]); -   $X + +; $y++; -M=m-2; -N=n-2; the         } -  Wuyi         returnresult; the}

Reference:http://www.cnblogs.com/springfor/p/3887890.html

*spiral Matrix