deploy: 73793ca

Author: yanglbme

en/lc/240/index.html

@@ -81449,8 +81449,8 @@ <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
 <h3 id="solution-1-in-place-rotation">Solution 1: In-place Rotation</h3>
-<p>According to the problem requirements, we actually need to rotate $matrix[i][j]$ to $matrix[j][n - i - 1]$.</p>
-<p>We can first flip the matrix upside down, that is, swap $matrix[i][j]$ and $matrix[n - i - 1][j]$, and then flip the matrix along the main diagonal, that is, swap $matrix[i][j]$ and $matrix[j][i]$. This way, we can rotate $matrix[i][j]$ to $matrix[j][n - i - 1]$.</p>
+<p>According to the problem requirements, we need to rotate $\text{matrix}[i][j]$ to $\text{matrix}[j][n - i - 1]$.</p>
+<p>We can first flip the matrix upside down, i.e., swap $\text{matrix}[i][j]$ with $\text{matrix}[n - i - 1][j]$, and then flip the matrix along the main diagonal, i.e., swap $\text{matrix}[i][j]$ with $\text{matrix}[j][i]$. This way, we can rotate $\text{matrix}[i][j]$ to $\text{matrix}[j][n - i - 1]$.</p>
 <p>The time complexity is $O(n^2)$, where $n$ is the side length of the matrix. The space complexity is $O(1)$.</p>
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