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leetcode-cn/problem (Chinese)/砖墙 [brick-wall].html
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<p>给你一个二维数组 <code>wall</code> ，该数组包含这堵墙的相关信息。其中，<code>wall[i]</code> 是一个代表从左至右每块砖的宽度的数组。你需要找出怎样画才能使这条线 <strong>穿过的砖块数量最少</strong> ，并且返回 <strong>穿过的砖块数量</strong> 。</p>
<p> </p>
<p>&nbsp;</p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/cutwall-grid.jpg" style="width: 493px; height: 577px;" />
<img alt="" src="https://assets.leetcode.com/uploads/2025/01/17/a.png" style="width: 400px; height: 384px;" />
<pre>
<strong>输入：</strong>wall = [[1,2,2,1],[3,1,2],[1,3,2],[2,4],[3,1,2],[1,3,1,1]]
<strong>输出：</strong>2
@@ -19,15 +19,15 @@
<strong>输入：</strong>wall = [[1],[1],[1]]
<strong>输出：</strong>3
</pre>
 
&nbsp;
<p><strong>提示：</strong></p>
<ul>
<li><code>n == wall.length</code></li>
<li><code>1 <= n <= 10<sup>4</sup></code></li>
<li><code>1 <= wall[i].length <= 10<sup>4</sup></code></li>
<li><code>1 <= sum(wall[i].length) <= 2 * 10<sup>4</sup></code></li>
<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
<li><code>1 &lt;= wall[i].length &lt;= 10<sup>4</sup></code></li>
<li><code>1 &lt;= sum(wall[i].length) &lt;= 2 * 10<sup>4</sup></code></li>
<li>对于每一行 <code>i</code> ，<code>sum(wall[i])</code> 是相同的</li>
<li><code>1 <= wall[i][j] <= 2<sup>31</sup> - 1</code></li>
<li><code>1 &lt;= wall[i][j] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>