标签：
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix 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].
思路：
[ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ]
step 1: add top line- from left to right
step 2: add right line - from top to bottom
step 3: add bottom line - from right to left
step 4 : add left line - from bottom to top
注意边界条件: [2 3] 1.用总个数来限制，calling method会增加复杂度2.用 top bottom left right关系来限制
Solution 1:
public class Solution { public List<Integer> spiralOrder(int[][] matrix) { List<Integer> res=new ArrayList<Integer>(); if(matrix==null||matrix.length==0) { return res; } int l=matrix[0].length; int h=matrix.length; int total=l*h; int left=0,top=0,right=l-1,bottom=h-1; while(res.size()<total) { for(int i=left;i<=right;i++) { res.add(matrix[top][i]); } top++; if(res.size()<total) { for(int j=top;j<=bottom;j++) { res.add(matrix[j][right]); } right--; } if(res.size()<total) { for(int k=right;k>=left;k--) { res.add(matrix[bottom][k]); } bottom--; } if(res.size()<total) { for(int m=bottom;m>=top;m--) { res.add(matrix[m][left]); } left++; } } return res; } }
Solution2:
public class Solution { public List<Integer> spiralOrder(int[][] matrix) { List<Integer> res=new ArrayList<Integer>(); if(matrix==null||matrix.length==0) { return res; } int l=matrix[0].length; int h=matrix.length; int total=l*h; int left=0,top=0,right=l-1,bottom=h-1; while(res.size()<total) { for(int i=left;i<=right;i++) { res.add(matrix[top][i]); } top++; if(top>bottom) { break; } for(int j=top;j<=bottom;j++) { res.add(matrix[j][right]); } right--; if(right<left) { break; } for(int k=right;k>=left;k--) { res.add(matrix[bottom][k]); } bottom--; if(bottom<top) { break; } for(int m=bottom;m>=top;m--) { res.add(matrix[m][left]); } left++; } return res; } }
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原文地址：http://www.cnblogs.com/Machelsky/p/5898617.html
