deploy: 430ebdc

Author: yanglbme

en/lc/53/index.html

@@ -80364,17 +80364,17 @@ <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
 <h3 id="solution-1-dynamic-programming">Solution 1: Dynamic Programming</h3>
-<p>We define $f[i]$ to represent the maximum sum of the continuous subarray ending with the element $nums[i]$. Initially, $f[0] = nums[0]$. The final answer we are looking for is $\max_{0 \leq i &lt; n} f[i]$.</p>
-<p>Consider $f[i]$, where $i \geq 1$, its state transition equation is:</p>
+<p>We define $f[i]$ to represent the maximum sum of a contiguous subarray ending at element $\textit{nums}[i]$. Initially, $f[0] = \textit{nums}[0]$. The final answer we seek is $\max_{0 \leq i &lt; n} f[i]$.</p>
+<p>Consider $f[i]$ for $i \geq 1$. Its state transition equation is:</p>
 <p>$$
-f[i] = \max { f[i - 1] + nums[i], nums[i] }
+f[i] = \max(f[i - 1] + \textit{nums}[i], \textit{nums}[i])
 $$</p>
-<p>Which is also:</p>
+<p>That is:</p>
 <p>$$
-f[i] = \max { f[i - 1], 0 } + nums[i]
+f[i] = \max(f[i - 1], 0) + \textit{nums}[i]
 $$</p>
-<p>Since $f[i]$ is only related to $f[i - 1]$, we can use a single variable $f$ to maintain the current value of $f[i]$, and then perform state transition. The answer is $\max_{0 \leq i &lt; n} f$.</p>
-<p>The time complexity is $O(n)$, where $n$ is the length of the array $nums$. We only need to traverse the array once to get the answer. The space complexity is $O(1)$, we only need constant space to store several variables.</p>
+<p>Since $f[i]$ is only related to $f[i - 1]$, we can use a single variable $f$ to maintain the current value of $f[i]$ and perform the state transition. The answer is $\max_{0 \leq i &lt; n} f$.</p>
+<p>The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.</p>
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en/search/search_index.json

@@ -84967,17 +84967,17 @@ <h2 id="_2">解法</h2>
 <!-- solution:start -->
 
 <h3 id="_3">方法一：动态规划</h3>
-<p>我们定义 $f[i]$ 表示以元素 $nums[i]$ 为结尾的连续子数组的最大和，初始时 $f[0] = nums[0]$，那么最终我们要求的答案即为 $\max_{0 \leq i &lt; n} f[i]$。</p>
+<p>我们定义 $f[i]$ 表示以元素 $\textit{nums}[i]$ 为结尾的连续子数组的最大和，初始时 $f[0] = \textit{nums}[0]$，那么最终我们要求的答案即为 $\max_{0 \leq i &lt; n} f[i]$。</p>
 <p>考虑 $f[i]$，其中 $i \geq 1$，它的状态转移方程为：</p>
 <p>$$
-f[i] = \max { f[i - 1] + nums[i], nums[i] }
+f[i] = \max(f[i - 1] + \textit{nums}[i], \textit{nums}[i])
 $$</p>
 <p>也即：</p>
 <p>$$
-f[i] = \max { f[i - 1], 0 } + nums[i]
+f[i] = \max(f[i - 1], 0) + \textit{nums}[i]
 $$</p>
 <p>由于 $f[i]$ 只与 $f[i - 1]$ 有关系，因此我们可以只用一个变量 $f$ 来维护对于当前 $f[i]$ 的值是多少，然后进行状态转移即可。答案为 $\max_{0 \leq i &lt; n} f$。</p>
-<p>时间复杂度 $O(n)$，其中 $n$ 为数组 $nums$ 的长度。我们只需要遍历一遍数组即可求得答案。空间复杂度 $O(1)$，我们只需要常数空间存放若干变量。</p>
+<p>时间复杂度 $O(n)$，其中 $n$ 为数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。</p>
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