给你一个 m
行 n
列的矩阵 matrix
,请按照 顺时针螺旋顺序 ,返回矩阵中的所有元素。
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]
提示:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100
public class Solution {
public IList SpiralOrder(int[][] matrix) {
List ilist= new List();
int row = matrix.Length;
int col = matrix[0].Length;
bool[,] isUser = new bool[row,col];
int dir=1;
int m=0,n=0;
ilist.Add(matrix[m][n]);
isUser[0,0]=true;
while (true)
{
switch (dir)
{
case 1:
if (n + 1 >= col) dir = 2;
break;
case 2:
if (m + 1 >= row) dir = 3;
break;
case 3:
if (n - 1 < 0) dir = 4;
break;
case 4:
if (m - 1 < 0) dir = 1;
break;
default:
break;
}
switch (dir)
{
case 1:
if(n + 1 =0)
if (isUser[m, n -1]) dir = 4;
break;
case 4:
if(m-1>=0)
if (isUser[m - 1, n]) dir = 1;
break;
default:
break;
}
switch (dir)
{
case 1:
if(n + 1 =0){
if (isUser[m, n - 1]) return ilist;
n--;
}
else return ilist;
break;
case 4:
if(m-1>=0){
if (isUser[m - 1, n]) return ilist;
m--;
}
else return ilist;
break;
default:
break;
}
isUser[m, n] = true;
ilist.Add(matrix[m][n]);
}
return ilist;
}
}