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Commit 430ebdc

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fix: update solutions to lc problem: No.53 (#3740)
close #3734 
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‎solution/0000-0099/0053.Maximum Subarray/README.md‎

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### 方法一:动态规划
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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],ドル那么最终我们要求的答案即为 $\max_{0 \leq i < n} f[i]$。
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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],ドル那么最终我们要求的答案即为 $\max_{0 \leq i < n} f[i]$。
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考虑 $f[i],ドル其中 $i \geq 1,ドル它的状态转移方程为:
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$$
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f[i] = \max \{ f[i - 1] + nums[i], nums[i] \}
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f[i] = \max(f[i - 1] + \textit{nums}[i], \textit{nums}[i])
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$$
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也即:
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$$
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f[i] = \max \{ f[i - 1], 0 \} + nums[i]
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f[i] = \max(f[i - 1], 0) + \textit{nums}[i]
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$$
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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 $\max_{0 \leq i < n} f$。
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时间复杂度 $O(n),ドル其中 $n$ 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。
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时间复杂度 $O(n),ドル其中 $n$ 为数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。
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‎solution/0000-0099/0053.Maximum Subarray/README_EN.md‎

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### Solution 1: Dynamic Programming
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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 $\max_{0 \leq i < n} f[i]$.
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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 $\max_{0 \leq i < n} f[i]$.
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Consider $f[i]$, where $i \geq 1$, its state transition equation is:
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Consider $f[i]$ for $i \geq 1$. Its state transition equation is:
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$$
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f[i] = \max \{ f[i - 1] + nums[i], nums[i] \}
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f[i] = \max(f[i - 1] + \textit{nums}[i], \textit{nums}[i])
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$$
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Which is also:
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That is:
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$$
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f[i] = \max \{ f[i - 1], 0 \} + nums[i]
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f[i] = \max(f[i - 1], 0) + \textit{nums}[i]
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$$
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Since $f[i]$ is only related to $f[i - 1],ドル we can use a single variable $f$ to maintain the current value of $f[i]$, and then perform state transition. The answer is $\max_{0 \leq i < n} f$.
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Since $f[i]$ is only related to $f[i - 1],ドル we can use a single variable $f$ to maintain the current value of $f[i]$ and perform the state transition. The answer is $\max_{0 \leq i < n} f$.
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The time complexity is $O(n),ドル 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.
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The time complexity is $O(n),ドル where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
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