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## Problem

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]


Example 2:

Input:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]


## Code

class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<Integer>();

if (matrix.length == 0) {
return res;
}

int rowBegin = 0;
int rowEnd = matrix.length-1;
int colBegin = 0;
int colEnd = matrix[0].length - 1;

while (rowBegin <= rowEnd && colBegin <= colEnd) {
// Traverse Right
for (int j = colBegin; j <= colEnd; j++) {
res.add(matrix[rowBegin][j]);
}
rowBegin++;

// Traverse Down
for (int j = rowBegin; j <= rowEnd; j++) {
res.add(matrix[j][colEnd]);
}
colEnd--;

if (rowBegin <= rowEnd) {
// Traverse Left
for (int j = colEnd; j >= colBegin; j--) {
res.add(matrix[rowEnd][j]);
}
}
rowEnd--;

if (colBegin <= colEnd) {
// Traverse Up
for (int j = rowEnd; j >= rowBegin; j--) {
res.add(matrix[j][colBegin]);
}
}
colBegin++;
}

return res;
}
}