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.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
从外往里一圈一圈遍历并存储矩阵元素即可。
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
def add(i1, j1, i2, j2):
if i1 == i2:
return [matrix[i1][j] for j in range(j1, j2 + 1)]
if j1 == j2:
return [matrix[i][j1] for i in range(i1, i2 + 1)]
return [matrix[i1][j] for j in range(j1, j2)] + [matrix[i][j2] for i in range(i1, i2)] + [matrix[i2][j] for j in range(j2, j1, -1)] + [matrix[i][j1] for i in range(i2, i1, -1)]
m, n = len(matrix), len(matrix[0])
i1, j1, i2, j2 = 0, 0, m - 1, n - 1
res = []
while i1 <= i2 and j1 <= j2:
res += add(i1, j1, i2, j2)
i1, j1, i2, j2 = i1 + 1, j1 + 1, i2 - 1, j2 - 1
return resclass Solution {
private List<Integer> res;
public List<Integer> spiralOrder(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
res = new ArrayList<>();
int i1 = 0, i2 = m - 1;
int j1 = 0, j2 = n - 1;
while (i1 <= i2 && j1 <= j2) {
add(matrix, i1++, j1++, i2--, j2--);
}
return res;
}
private void add(int[][] matrix, int i1, int j1, int i2, int j2) {
if (i1 == i2) {
for (int j = j1; j <= j2; ++j) {
res.add(matrix[i1][j]);
}
return;
}
if (j1 == j2) {
for (int i = i1; i <= i2; ++i) {
res.add(matrix[i][j1]);
}
return;
}
for (int j = j1; j < j2; ++j) {
res.add(matrix[i1][j]);
}
for (int i = i1; i < i2; ++i) {
res.add(matrix[i][j2]);
}
for (int j = j2; j > j1; --j) {
res.add(matrix[i2][j]);
}
for (int i = i2; i > i1; --i) {
res.add(matrix[i][j1]);
}
}
}