def spiral_traversal(matrix): res = [] if len(matrix) == 0: return res row_begin = 0 row_end = len(matrix) - 1 col_begin = 0 col_end = len(matrix[0]) - 1 while row_begin <= row_end and col_begin <= col_end: for i in range(col_begin, col_end+1): res.append(matr...
1publicList<Integer>spiralOrder(int[][]matrix){2List<Integer>res=newArrayList<>();34if(matrix==null||matrix.length==0||matrix[0].length==0){5returnres;6}78this.printSpiralOrder(0,0,matrix.length,matrix[0].length,res,matrix);910returnres;11}1213// printSpiralOrder is a poor name, f...
Given a 2d array of n*n and the task is to find the antispiral arrangement of the given matrix Input : arr[4][4]={1,2,3,4, 5,6,7,8, 9,10,11,12 13,14,15,16} Output: 10 11 7 6 5 9 13 14 15 16 12 8 4 3 2 1 For this, stack can be used where the transpose ...