Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
Example 1:
Input: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] Output: [1,2,3,4,8,12,11,10,9,5,6,7]
题意:
蛇形输出。
思路:
不同于之前的蛇形填数,此题的矩阵可以是长方形的。但可以类比于之前的,设置top,left,right,down四个参数,模拟蛇形进行输出,注意判断截止条件,当top==down||left==right时还可以继续添加,比如3*3的矩阵中,top=down时,num[1][0],num[1][1]尚未添加。所以截止条件是top==down+1||left==right+1.
代码:
class Solution { public List<Integer> spiralOrder(int[][] matrix) { List list=new ArrayList(); if(matrix.length==0) return list; int left=0,top=0,right=matrix[0].length-1,down=matrix.length-1; int t=0; while(left<=right||top<=down) { for(int i=left;i<=right;i++) list.add(matrix[top][i]); top++; if(top==down+1) break; for(int i=top;i<=down;i++) list.add(matrix[i][right]); right--; if(left==right+1) break; for(int i=right;i>=left;i--) list.add(matrix[down][i]); down--; if(top==down+1) break; for(int i=down;i>=top;i--) list.add(matrix[i][left]); left++; if(left==right+1) break; } return list; } }