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% Most of the following are not required for the majority
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% Document Info
\author{woshiamiaojiang}
\pdfinfo{
/Title (labuladong.pdf)
/Creator (Cheatography)
/Author (woshiamiaojiang)
/Subject (Labuladong Cheat Sheet)
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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\vspace{-7pt}
{\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
\hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Labuladong Cheat Sheet}}}} \\
\normalsize{by \textcolor{DarkBackground}{woshiamiaojiang} via \textcolor{DarkBackground}{\uline{cheatography.com/193748/cs/40865/}}}
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\noindent
\begin{multicols}{3}
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\mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}} \\
\vspace{-2pt}woshiamiaojiang \\
\uline{cheatography.com/woshiamiaojiang} \\
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\mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}} \\
\vspace{-2pt}Not Yet Published.\\
Updated 17th October, 2023.\\
Page {\thepage} of \pageref{LastPage}.
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\vfill
\columnbreak
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\mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}} \\
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\vspace{-5pt}
%\includegraphics[width=48px,height=48px]{dave.jpeg}
Measure your website readability!\\
www.readability-score.com
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\begin{tabularx}{17.67cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{17.67cm}}{\bf\textcolor{white}{124. \seqsplit{二叉树中的最大路径和}}} \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{17.67cm}}{class Solution \{ \newline \newline int res = Integer.MIN\_VALUE; \newline \newline public int maxPathSum(TreeNode root) \{ \newline getOneSideMax(root); \newline return res; \newline \} \newline \newline public int getOneSideMax(TreeNode root) \{ \newline if (root == null) \{ \newline return 0; \newline \} \newline int left = Math.max(0, \seqsplit{getOneSideMax(root.left));} \newline int right = Math.max(0, \seqsplit{getOneSideMax(root.right));} \newline res = Math.max(left + right + root.val, res); \newline return Math.max(left, right) + root.val; \newline \} \newline \}} \tn
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\mymulticolumn{1}{x{17.67cm}}{\bf\textcolor{white}{230. \seqsplit{二叉搜索树中第K小的元素}}} \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{17.67cm}}{class Solution \{ \newline \newline int k = 0; \newline int res = 0; \newline \newline public int kthSmallest(TreeNode root, int k) \{ \newline this.k = k; \newline traverse(root); \newline return res; \newline \} \newline \newline public void traverse(TreeNode root) \{ \newline if (root == null) \{ \newline return; \newline \} \newline traverse(root.left); \newline k-{}-; \newline if (k == 0) \{ \newline res = root.val; \newline return; \newline \} \newline traverse(root.right); \newline \} \newline \}} \tn
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\begin{tabularx}{17.67cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{17.67cm}}{\bf\textcolor{white}{322. 零钱兑换}} \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{17.67cm}}{/{\emph{ \newline 开辟一个动态数组，基底dp{[}0{]}=0。 \newline dp{[}i{]} = num。 \newline \seqsplit{i代表目标值，num代表硬币个数。} \newline \seqsplit{num初值为amt+1。代表不可能。} \newline 每个dp{[}i{]}都与dp{[}i-coin{]}的结果相关联。 \newline }}/ \newline class Solution \{ \newline public int coinChange(int{[}{]} coins, int amount) \{ \newline \newline int{[}{]} dp = new int{[}amount + 1{]}; \newline Arrays.fill(dp, amount + 1); \newline dp{[}0{]} = 0; \newline for (int i = 1; i \textless{} dp.length; i++) \{ \newline for (int coin : coins) \{ \newline if (i - coin \textgreater{}= 0) \{ \newline dp{[}i{]} = Math.min(dp{[}i - coin{]} + 1, dp{[}i{]}); \newline \} \newline \} \newline \} \newline return dp{[}dp.length - 1{]} == amount + 1? -1 : dp{[}dp.length - 1{]}; \newline \} \newline \}} \tn
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\begin{tabularx}{17.67cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{17.67cm}}{\bf\textcolor{white}{46. 全排列}} \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{17.67cm}}{class Solution \{ \newline \newline boolean{[}{]} used; \newline List\textless{}List\textless{}Integer\textgreater{}\textgreater{} res; \newline \newline public List\textless{}List\textless{}Integer\textgreater{}\textgreater{} permute(int{[}{]} nums) \{ \newline res = new LinkedList\textless{}\textgreater{}(); \newline used = new boolean{[}nums.length{]}; \newline backtrack(nums, new LinkedList\textless{}\textgreater{}()); \newline return res; \newline \} \newline \newline public void backtrack(int{[}{]} nums, LinkedList\textless{}Integer\textgreater{} track) \{ \newline // \seqsplit{临时track大小满的时候} \newline if (track.size() == nums.length) \{ \newline res.add(new LinkedList(track)); \newline return; \newline \} \newline for (int i = 0; i \textless{} nums.length; i++) \{ \newline if (used{[}i{]}) continue; \newline // \seqsplit{做选择，添加，下探结果} \newline used{[}i{]} = true; \newline track.add(nums{[}i{]}); \newline backtrack(nums, track); \newline used{[}i{]} = false; \newline track.removeLast(); \newline \} \newline \} \newline \}} \tn
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\begin{tabularx}{17.67cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{17.67cm}}{\bf\textcolor{white}{105. \seqsplit{从前序与中序遍历序列构造二叉树}}} \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{17.67cm}}{class Solution \{ \newline \newline \newline HashMap\textless{}Integer, Integer\textgreater{} inorderValMapIdx = new HashMap\textless{}\textgreater{}(); \newline \newline public TreeNode buildTree(int{[}{]} preorder, int{[}{]} inorder) \{ \newline for (int i = 0; i \textless{} inorder.length; i++) \{ \newline inorderValMapIdx.put(inorder{[}i{]}, i); \newline \} \newline return buildTree(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1); \newline \} \newline \newline // pre: 3 9 20 15 7 \newline // T L R \newline // in : 9 3 15 20 7 \newline // L T R \newline public TreeNode buildTree(int{[}{]} preorder, int{[}{]} inorder, int preStart, int preEnd, int inStart, int inEnd) \{ \newline if (preStart \textgreater{} preEnd) \{ \newline return null; \newline \} \newline int rootVal = preorder{[}preStart{]}; \newline int inorderRootIdx = \seqsplit{inorderValMapIdx.get(rootVal);} \newline TreeNode root = new TreeNode(rootVal); \newline root.left = buildTree(preorder, inorder, preStart + 1, preStart + inorderRootIdx - inStart, inStart, inorderRootIdx - 1); \newline root.right = buildTree(preorder, inorder, preStart + inorderRootIdx - inStart + 1, preEnd, inorderRootIdx + 1, inEnd); \newline return root; \newline \} \newline \}} \tn
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