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Description
tab: English
<p>Given an <code>m x n</code> <code>matrix</code>, return <em>all elements of the</em> <code>matrix</code> <em>in spiral order</em>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [1,2,3,6,9,8,7,4,5]
</pre>
<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
<strong>Output:</strong> [1,2,3,4,8,12,11,10,9,5,6,7]
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>m == matrix.length</code></li>
<li><code>n == matrix[i].length</code></li>
<li><code>1 <= m, n <= 10</code></li>
<li><code>-100 <= matrix[i][j] <= 100</code></li>
</ul>
> [!tip]- Hint 1
>
> Well for some problems, the best way really is to come up with some algorithms for simulation. Basically, you need to simulate what the problem asks us to do.
> [!tip]- Hint 2
>
> We go boundary by boundary and move inwards. That is the essential operation. First row, last column, last row, first column, and then we move inwards by 1 and repeat. That's all. That is all the simulation that we need.
> [!tip]- Hint 3
>
> Think about when you want to switch the progress on one of the indexes. If you progress on i out of [i, j], you'll shift in the same column. Similarly, by changing values for j, you'd be shifting in the same row.
Also, keep track of the end of a boundary so that you can move inwards and then keep repeating. It's always best to simulate edge cases like a single column or a single row to see if anything breaks or not.
---
[submissions](https://leetcode.com/problems/spiral-matrix/submissions/) | [solutions](https://leetcode.com/problems/spiral-matrix/solutions/)
tab: 中文
<p>给你一个 <code>m</code> 行 <code>n</code> 列的矩阵 <code>matrix</code> ，请按照 <strong>顺时针螺旋顺序</strong> ，返回矩阵中的所有元素。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>输入：</strong>matrix = [[1,2,3],[4,5,6],[7,8,9]]
<strong>输出：</strong>[1,2,3,6,9,8,7,4,5]
</pre>
<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>输入：</strong>matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
<strong>输出：</strong>[1,2,3,4,8,12,11,10,9,5,6,7]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>m == matrix.length</code></li>
<li><code>n == matrix[i].length</code></li>
<li><code>1 <= m, n <= 10</code></li>
<li><code>-100 <= matrix[i][j] <= 100</code></li>
</ul>
> [!tip]- 提示 1
>
> Well for some problems, the best way really is to come up with some algorithms for simulation. Basically, you need to simulate what the problem asks us to do.
> [!tip]- 提示 2
>
> We go boundary by boundary and move inwards. That is the essential operation. First row, last column, last row, first column, and then we move inwards by 1 and repeat. That's all. That is all the simulation that we need.
> [!tip]- 提示 3
>
> Think about when you want to switch the progress on one of the indexes. If you progress on i out of [i, j], you'll shift in the same column. Similarly, by changing values for j, you'd be shifting in the same row.
Also, keep track of the end of a boundary so that you can move inwards and then keep repeating. It's always best to simulate edge cases like a single column or a single row to see if anything breaks or not.
---
[提交记录](https://leetcode.cn/problems/spiral-matrix/submissions/) | [题解](https://leetcode.cn/problems/spiral-matrix/solution/)
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