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Commit b2a6b78

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feat: add solutions to lc problems: No.0328~0330 (doocs#2265)
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‎solution/0300-0399/0328.Odd Even Linked List/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Single Pass
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We can use two pointers $a$ and $b$ to represent the tail nodes of the odd and even nodes respectively. Initially, pointer $a$ points to the head node $head$ of the list, and pointer $b$ points to the second node $head.next$ of the list. In addition, we use a pointer $c$ to point to the head node $head.next$ of the even nodes, which is the initial position of pointer $b$.
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We traverse the list, set pointer $a$ to point to the next node of $b,ドル i.e., $a.next = b.next,ドル then move pointer $a$ back by one position, i.e., $a = a.next$; set pointer $b$ to point to the next node of $a,ドル i.e., $b.next = a.next,ドル then move pointer $b$ back by one position, i.e., $b = b.next$. Continue to traverse until $b$ reaches the end of the list.
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Finally, we set the tail node $a$ of the odd nodes to point to the head node $c$ of the even nodes, i.e., $a.next = c,ドル then return the head node $head$ of the list.
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The time complexity is $O(n),ドル where $n$ is the length of the list, and we need to traverse the list once. The space complexity is $O(1)$. We only need to maintain a limited number of pointers.
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‎solution/0300-0399/0329.Longest Increasing Path in a Matrix/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Memoization Search
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We design a function $dfs(i, j),ドル which represents the length of the longest increasing path that can be obtained starting from the coordinate $(i, j)$ in the matrix. The answer is $\max_{i, j} \textit{dfs}(i, j)$.
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The execution logic of the function $dfs(i, j)$ is as follows:
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- If $(i, j)$ has been visited, directly return $\textit{f}(i, j)$;
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- Otherwise, search $(i, j),ドル search the coordinates $(x, y)$ in four directions. If 0ドル \le x < m, 0 \le y < n$ and $matrix[x][y] > matrix[i][j],ドル then search $(x, y)$. After the search is over, update $\textit{f}(i, j)$ to $\textit{f}(i, j) = \max(\textit{f}(i, j), \textit{f}(x, y) + 1)$. Finally, return $\textit{f}(i, j)$.
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The time complexity is $O(m \times n),ドル and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the matrix, respectively.
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Similar problems:
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- [2328. Number of Increasing Paths in a Grid](https://github.com/doocs/leetcode/blob/main/solution/2300-2399/2328.Number%20of%20Increasing%20Paths%20in%20a%20Grid/README_EN.md)
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‎solution/0300-0399/0330.Patching Array/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Greedy
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Let's assume that the number $x$ is the smallest positive integer that cannot be represented. Then all the numbers in $[1,..x-1]$ can be represented. In order to represent the number $x,ドル we need to add a number that is less than or equal to $x$:
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- If the added number equals $x,ドル since all numbers in $[1,..x-1]$ can be represented, after adding $x,ドル all numbers in the range $[1,..2x-1]$ can be represented, and the smallest positive integer that cannot be represented becomes 2ドルx$.
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- If the added number is less than $x,ドル let's assume it's $x',ドル since all numbers in $[1,..x-1]$ can be represented, after adding $x',ドル all numbers in the range $[1,..x+x'-1]$ can be represented, and the smallest positive integer that cannot be represented becomes $x+x' \lt 2x$.
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Therefore, we should greedily add the number $x$ to cover a larger range.
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We use a variable $x$ to record the current smallest positive integer that cannot be represented, initialized to 1ドル$. At this time, $[1,..x-1]$ is empty, indicating that no number can be covered; we use a variable $i$ to record the current index of the array being traversed.
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We perform the following operations in a loop:
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- If $i$ is within the range of the array and $nums[i] \le x,ドル it means that the current number can be covered, so we add the value of $nums[i]$ to $x,ドル and increment $i$ by 1ドル$.
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- Otherwise, it means that $x$ is not covered, so we need to supplement a number $x$ in the array, and then update $x$ to 2ドルx$.
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- Repeat the above operations until the value of $x$ is greater than $n$.
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The final answer is the number of supplemented numbers.
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The time complexity is $O(m + \log n),ドル where $m$ is the length of the array $nums$. The space complexity is $O(1)$.
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‎solution/util.py‎

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category_readme_en = load_template("category_readme_template_en")
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category_dict = {
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'Database': '数据库',
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"Database": "数据库",
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}
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