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].
此题关键点在于开始要设置四个边节点，在设置一个遍历变量（分别依次代表col，row，col，row）。接下来出界条件是begin>end值，然后每次遍历一行或一列的时候，把那一行或那一列的边界值去掉，代码如下：
public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<Integer>();
if(matrix==null||matrix.length==0||matrix[0].length==0) return res;
int rowbegin = 0;
int rowend = matrix.length-1;
int colbegin = 0;
int colend = matrix[0].length-1;
while(true){
for(int i=colbegin;i<=colend;i++){
res.add(matrix[rowbegin][i]);
}
rowbegin++;
if(rowbegin>rowend) break;
for(int i=rowbegin;i<=rowend;i++){
res.add(matrix[i][colend]);
}
colend--;
if(colbegin>colend) break;
for(int i=colend;i>=colbegin;i--){
res.add(matrix[rowend][i]);
}
rowend--;
if(rowbegin>rowend) break;
for(int i=rowend;i>=rowbegin;i--){
res.add(matrix[i][colbegin]);
}
colbegin++;
if(colbegin>colend) break;
}
return res;
}
}
原文：http://www.cnblogs.com/codeskiller/p/6363918.html