Lintcode-medium-spiral Matrix

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

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] .

 Public classSolution {/**     * @parammatrix A matrix of M x n elements *@returnAn integer list*/     PublicList<integer> Spiralorder (int[] matrix) {        //Write Your code hereList<Integer> res =NewArraylist<integer>(); if(Matrix = =NULL|| Matrix.length = = 0 | | Matrix[0] = =NULL|| Matrix[0].length = = 0)            returnRes; intm =matrix.length; intn = matrix[0].length; intLayer = 0;  while(Layer < M-LAYER-1) && (Layer < n-1-layer)) {                         for(inti = 0; I < n-layer-1-layer; i++) Res.add (Matrix[layer][layer+i]);  for(inti = 0; I < m-layer-1-layer; i++) Res.add (Matrix[layer+ i][n-layer-1]);  for(inti = 0; I < n-layer-1-layer; i++) Res.add (matrix[m-Layer-1][n-layer-1-i]);  for(inti = 0; I < m-layer-1-layer; i++) Res.add (matrix[m-Layer-1-i]                        [Layer]); Layer++; }                if(M >N) {            if(Layer = = n-1-layer) {                 for(inti = layer; I <= m-layer-1; i++) Res.add (Matrix[i][layer]); }        }        Else if(M <N) {            if(Layer = = m-1-layer) {                 for(inti = layer; I <= n-layer-1; i++) Res.add (Matrix[layer][i]); }        }        Else{            if(Layer = = m-1-layer)            {Res.add (Matrix[layer][layer]); }        }                returnRes; }    }

Lintcode-medium-spiral Matrix