输入一个矩阵,按照从外向里以顺时针的顺序依次打印出每一个数字。
示例 1:
输入:matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出:[1,2,3,6,9,8,7,4,5]
示例 2:
输入:matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出:[1,2,3,4,8,12,11,10,9,5,6,7]
限制:
0 <= matrix.length <= 100
0 <= matrix[i].length <= 100
代码:
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;
}
}