deploy: 355400e

Author: yanglbme

en/lc/559/index.html

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+  <a href="#solution-1-dynamic-programming" class="md-nav__link">
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-      Solution 1
+      Solution 1: Dynamic Programming
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@@ -81910,7 +81910,13 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
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-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-dynamic-programming">Solution 1: Dynamic Programming</h3>
+<p>We define $f[i][j][k]$ to represent the length of the longest consecutive $1$s ending at $(i, j)$ in direction $k$. The value range of $k$ is $0, 1, 2, 3$, representing horizontal, vertical, diagonal, and anti-diagonal directions, respectively.</p>
+<blockquote>
+<p>We can also use four 2D arrays to represent the length of the longest consecutive $1$s in the four directions.</p>
+</blockquote>
+<p>We traverse the matrix, and when we encounter $1$, we update the value of $f[i][j][k]$. For each position $(i, j)$, we only need to update the values in its four directions. Then we update the answer.</p>
+<p>The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns in the matrix, respectively.</p>
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en/lc/562/index.html

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       <ul class="md-nav__list">
 
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-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-simulation" class="md-nav__link">
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-      Solution 1
+      Solution 1: Simulation
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 <h2 id="solutions">Solutions</h2>
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-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-simulation">Solution 1: Simulation</h3>
+<p>First, we get the number of rows and columns of the original matrix, denoted as $m$ and $n$ respectively. If $m \times n \neq r \times c$, then the matrix cannot be reshaped, and we return the original matrix directly.</p>
+<p>Otherwise, we create a new matrix with $r$ rows and $c$ columns. Starting from the first element of the original matrix, we traverse all elements in row-major order and place the traversed elements into the new matrix in order.</p>
+<p>After traversing all elements of the original matrix, we get the answer.</p>
+<p>The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the original matrix, respectively. Ignoring the space consumption of the answer, the space complexity is $O(1)$.</p>
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en/lc/563/index.html

@@ -86522,7 +86522,7 @@ <h3 id="_3">方法一：动态规划</h3>
 <p>我们也可以用四个二维数组分别表示四个方向的最长连续 $1$ 的长度。</p>
 </blockquote>
 <p>遍历矩阵，当遇到 $1$ 时，更新 $f[i][j][k]$ 的值。对于每个位置 $(i, j)$，我们只需要更新其四个方向的值即可。然后更新答案。</p>
-<p>时间复杂度 $O(m\times n)$，空间复杂度 $O(m\times n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。</p>
+<p>时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="1:4"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label></div>
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