輸入一個矩陣，按照從外向里以順時針的順序依次打印出每一個數字。

?

示例 1：

輸入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
輸出：[1,2,3,6,9,8,7,4,5]
示例 2：

輸入：matrix =?[[1,2,3,4],[5,6,7,8],[9,10,11,12]]
輸出：[1,2,3,4,8,12,11,10,9,5,6,7]
?

限制：

0 <= matrix.length <= 100
0 <= matrix[i].length?<= 100

代碼：

class?Solution?{

????public?int[]?spiralOrder(int[][]?matrix)?{

???????

????????int?m?=?matrix.length;

????????if(m==0)

????????{

????????????int?nums[]?=?new?int[m];

????????????return?nums;

????????}

????????int?n?=?matrix[0].length;

????????int?nums[]?=?new?int[m*n];

?????????if(matrix.length==0)

????????{

????????????return?nums;

????????}

????????

????????int?k?=0,count=0;

????????while(count<m*n)

????????{

????????????for(int?i=k;i<matrix[0].length-k;i++)

????????????{

????????????????nums[count++]=matrix[k][i];

????????????}

????????????if(count==m*n)?break;

????????????for(int?i=k+1;i<matrix.length-k;i++)

????????????{

????????????????nums[count++]?=?matrix[i][matrix[0].length-k-1];

????????????}

????????????if(count==m*n)?break;

????????????for(int?i=matrix[0].length-k-2;i>=k;i--)

????????????{

????????????????nums[count++]?=?matrix[matrix.length-1-k][i];

????????????}

????????????if(count==m*n)?break;

????????????for(int?i=matrix.length-2-k;i>k;i--)

????????????{

????????????????nums[count++]?=?matrix[i][k];

????????????}

????????????if(count==m*n)?break;

????????????k++;

????????}

????????return?nums;

????}

}

總結

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