顺时针打印m*n矩阵

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

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].

package com.vic.leetcode.solutions;

import java.util.ArrayList;
import java.util.List;

public class Solution {
	public List<Integer> spiralOrder(int[][] matrix) {
		List<Integer> list = new ArrayList<Integer>();
		if(null==matrix||matrix.length==0)
			return list;
		int start = 0,columns = matrix[0].length,rows = matrix.length;
		while(columns>2*start&&rows>2*start){
			 list.addAll(spiralPrint(matrix,columns,rows,start));
			 start++;
		}
		return list;
	}
	
	private List<Integer> spiralPrint(int[][] matrix,int columns, int rows, int start) {
		List<Integer> tempList = new ArrayList<Integer>();
		int endX = columns - start - 1;
		int endY = rows - start - 1;
		//from left to right
		for(int i = start; i <= endX; i++){
			tempList.add(matrix[start][i]);
		}
		//from top to bottom
		if(endY > start){
			for(int i = start + 1; i <= endY; i++){
				tempList.add(matrix[i][endX]);
			}
		}
		//from right to left
		if(endX > start && endY > start){
			for(int i = endX - 1; i >= start; i--){
				tempList.add(matrix[endY][i]);
			}
		}
		//from bottom to top
		if(endX > start && endY > start){
			for(int i = endY - 1; i > start; i--){
				tempList.add(matrix[i][start]);
			}
		}
		return tempList;
	}

	public static void main(String[] args) {
		int[][] matrix = {{1,2,3},{4,5,6}};
		Solution sl = new Solution();
		sl.spiralOrder(matrix);
		for(Integer n : sl.spiralOrder(matrix)){
			System.out.print(n+" ");
		}
	}
}