## Spiral Matrix

### 描述

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

For example,
Given the following matrix:

```
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
```

You should return `[1,2,3,6,9,8,7,4,5]`.

### 分析

### 解法1 迭代

{% if book.java %}
```java
// Spiral Matrix
// 时间复杂度O(n^2)，空间复杂度O(1)
public class Solution {
public List spiralOrder(int[][] matrix) {
List result = new ArrayList<>();
if (matrix.length == 0) return result;
int beginX = 0, endX = matrix[0].length - 1;
int beginY = 0, endY = matrix.length - 1;
while (true) {
// From left to right
for (int j = beginX; j <= endX; ++j) result.add(matrix[beginY][j]);
if (++beginY > endY) break;
// From top to bottom
for (int i = beginY; i <= endY; ++i) result.add(matrix[i][endX]);
if (beginX > --endX) break;
// From right to left
for (int j = endX; j >= beginX; --j) result.add(matrix[endY][j]);
if (beginY > --endY) break;
// From bottom to top
for (int i = endY; i >= beginY; --i) result.add(matrix[i][beginX]);
if (++beginX > endX) break;
}
return result;
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// LeetCode, Spiral Matrix
// @author 龚陆安 (http://weibo.com/luangong)
// 时间复杂度O(n^2)，空间复杂度O(1)
class Solution {
public:
vector spiralOrder(vector >& matrix) {
vector result;
if (matrix.empty()) return result;
int beginX = 0, endX = matrix[0].size() - 1;
int beginY = 0, endY = matrix.size() - 1;
while (true) {
// From left to right
for (int j = beginX; j <= endX; ++j) result.push_back(matrix[beginY][j]);
if (++beginY > endY) break;
// From top to bottom
for (int i = beginY; i <= endY; ++i) result.push_back(matrix[i][endX]);
if (beginX > --endX) break;
// From right to left
for (int j = endX; j >= beginX; --j) result.push_back(matrix[endY][j]);
if (beginY > --endY) break;
// From bottom to top
for (int i = endY; i >= beginY; --i) result.push_back(matrix[i][beginX]);
if (++beginX > endX) break;
}
return result;
}
};
```
{% endif %}

### 解法2 递归

{% if book.java %}
```java
// Spiral Matrix
// 时间复杂度O(n^2)，空间复杂度O(1)
public class Solution {
public List spiralOrder(int[][] matrix) {
List result = new ArrayList<>();
if (matrix.length == 0) return result;

left = 0;
right = matrix[0].length - 1;
up = 0;
down = matrix.length - 1;
dfs(matrix, 0, 0, 0, result);
return result;
}

private void dfs(int[][] matrix, int i, int j, int direction,
List result) {
if (i < up || i > down) return;
if (j < left || j > right) return;

int nextDirection = (direction + 1) % 4;
switch (direction) {
case 0: // right
if (j < right) {
dfs(matrix, i, j + 1, direction, result);
} else {
++up;
dfs(matrix, i + 1, j, nextDirection, result);
}
break;
case 1: // down
if (i < down) {
dfs(matrix, i+1, j, direction, result);
} else {
--right;
dfs(matrix, i, j - 1, nextDirection, result);
}
break;
case 2: // left
if (j > left) {
dfs(matrix, i, j - 1, direction, result);
} else {
--down;
dfs(matrix, i - 1, j, nextDirection, result);
}
break;
default: // up
if (i > up) {
dfs(matrix, i - 1, j, direction, result);
} else {
++left;
dfs(matrix, i, j + 1, nextDirection, result);
}
}

}

private int left;
private int right;
private int up;
private int down;
}
```
{% endif %}

### 相关题目

* [Spiral Matrix II](spiral-matrix-ii.md)