输入一个矩阵 将这个矩阵顺时针以顺时针的顺序依次打印 java实现

/**

• 输入一个矩阵 将这个矩阵顺时针以顺时针的顺序依次打印

• 1 2 3

• 4 5 6 ——>1 2 3 6 9 8 7 4 5

• 7 8 9

• @author Administrator
  *解题的思路就是从最外层开始输出，然后一圈一圈的向内收缩输出
  */
  public class JuZhen {
  public static void main(String[] args) {
  //测试数据
  int [][]nums={{1,2,3},{4,5,6},{7,8,9}};
  int [][]nums2={{1,2,3,4},{5,6,7,8},{9,10,11,12}};
  int [][]nums3={{1,2},{3,4}};
  int [][]nums4={{1}};
  int [][]nums5={{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}};
  int [][]nums6={{1,2,3,4,5},{6,7,8,9,10},{11,12,13,14,15}};
  shuChu(0,0,nums6.length-1,nums6[0].length-1,nums6);
  }

  public static void shuChu(int starX,int starY,int endX,int endY,int [][]nums){

   //解决类似 7 8 9 这样的数 如果不写这个数组就会输出 7 8 9 8 7
   if(starX==endX){
   	//输出一层
   	for (int i = starY; i < endY+1; i++) {
   		System.out.print(nums[starX][i]+" ");
   	}
   	return;
   }
   if(starY==endY){
   	for (int i = starX+1; i < endX+1; i++) {
   		System.out.print(nums[i][endY]+" ");
   	}
   }
   
   //输出一层
   for (int i = starY; i < endY+1; i++) {
   	System.out.print(nums[starX][i]+" ");
   }	
   for (int i = starX+1; i < endX+1; i++) {
   	System.out.print(nums[i][endY]+" ");
   }
   for (int i = endY-1; i >=starY; i--) {
   	System.out.print(nums[endX][i]+" ");
   }
   for (int i = endX-1; i >starX; i--) {
   	System.out.print(nums[i][starY]+" ");
   }
   
   //判断此时这一个环的形态
   if(starX+1>endX-1||starY+1>endY-1){
   	//最后一个矩阵是矩形 高大于2
   	return;
   }
   if(starX==endX&&starY==endY){
   	//最后一个矩阵是的单个
   	return;
   }
   if(starX+1==endX&&starY+1==endY){
   	//最后一个矩阵是田字格
   	return;
   }
   shuChu(starX+1, starY+1, endX-1, endY-1, nums);
  

  }

}