Merge pull request #832 from openset/develop

A: new

Author: Shuo

Diff for: README.md

@@ -1,3 +1,3 @@
 module github.com/openset/leetcode
 
-go 1.14
+go 1.15

Diff for: go.mod

@@ -11,30 +11,42 @@
 
 ## [721. Accounts Merge (Medium)](https://leetcode.com/problems/accounts-merge "账户合并")
 
-<p>Given a list <code>accounts</code>, each element <code>accounts[i]</code> is a list of strings, where the first element <code>accounts[i][0]</code> is a <i>name</i>, and the rest of the elements are <i>emails</i> representing emails of the account.</p>
+<p>Given a list of <code>accounts</code> where each element <code>accounts[i]</code> is a list of strings, where the first element <code>accounts[i][0]</code> is a name, and the rest of the elements are <strong>emails</strong> representing emails of the account.</p>
 
-<p>Now, we would like to merge these accounts.  Two accounts definitely belong to the same person if there is some email that is common to both accounts.  Note that even if two accounts have the same name, they may belong to different people as people could have the same name.  A person can have any number of accounts initially, but all of their accounts definitely have the same name.</p>
+<p>Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some common email to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.</p>
 
-<p>After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails <b>in sorted order</b>.  The accounts themselves can be returned in any order.</p>
+<p>After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails <strong>in sorted order</strong>. The accounts themselves can be returned in <strong>any order</strong>.</p>
 
-<p><b>Example 1:</b><br />
-<pre style="white-space: pre-wrap">
-<b>Input:</b> 
-accounts = [["John", "[email protected]", "[email protected]"], ["John", "[email protected]"], ["John", "[email protected]", "[email protected]"], ["Mary", "[email protected]"]]
-<b>Output:</b> [["John", '[email protected]', '[email protected]', '[email protected]'],  ["John", "[email protected]"], ["Mary", "[email protected]"]]
-<b>Explanation:</b> 
-The first and third John's are the same person as they have the common email "[email protected]".
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> accounts = [[&quot;John&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;John&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Mary&quot;,&quot;[email protected]&quot;],[&quot;John&quot;,&quot;[email protected]&quot;]]
+<strong>Output:</strong> [[&quot;John&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Mary&quot;,&quot;[email protected]&quot;],[&quot;John&quot;,&quot;[email protected]&quot;]]
+<strong>Explanation:</strong>
+The first and third John&#39;s are the same person as they have the common email &quot;[email protected]&quot;.
 The second John and Mary are different people as none of their email addresses are used by other accounts.
-We could return these lists in any order, for example the answer [['Mary', '[email protected]'], ['John', '[email protected]'], 
-['John', '[email protected]', '[email protected]', '[email protected]']] would still be accepted.
+We could return these lists in any order, for example the answer [[&#39;Mary&#39;, &#39;[email protected]&#39;], [&#39;John&#39;, &#39;[email protected]&#39;], 
+[&#39;John&#39;, &#39;[email protected]&#39;, &#39;[email protected]&#39;, &#39;[email protected]&#39;]] would still be accepted.
 </pre>
-</p>
 
-<p><b>Note:</b>
-<li>The length of <code>accounts</code> will be in the range <code>[1, 1000]</code>.</li>
-<li>The length of <code>accounts[i]</code> will be in the range <code>[1, 10]</code>.</li>
-<li>The length of <code>accounts[i][j]</code> will be in the range <code>[1, 30]</code>.</li>
-</p>
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> accounts = [[&quot;Gabe&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Kevin&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Ethan&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Hanzo&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Fern&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;]]
+<strong>Output:</strong> [[&quot;Ethan&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Gabe&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Hanzo&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Kevin&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;],[&quot;Fern&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;,&quot;[email protected]&quot;]]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= accounts.length &lt;= 1000</code></li>
+	<li><code>2 &lt;= accounts[i].length &lt;= 10</code></li>
+	<li><code>1 &lt;= accounts[i][j] &lt;= 30</code></li>
+	<li><code>accounts[i][0]</code> consists of English letters.</li>
+	<li><code>accounts[i][j] (for j &gt; 0)</code> is a valid email.</li>
+</ul>
 
 ### Related Topics
   [[Depth-first Search](../../tag/depth-first-search/README.md)]

Diff for: problems/accounts-merge/README.md

@@ -11,19 +11,36 @@
 
 ## [258. Add Digits (Easy)](https://leetcode.com/problems/add-digits "各位相加")
 
-<p>Given a non-negative integer <code>num</code>, repeatedly add all its digits until the result has only one digit.</p>
+<p>Given an integer <code>num</code>, repeatedly add all its digits until the result has only one digit, and return it.</p>
 
-<p><strong>Example:</strong></p>
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
 
 <pre>
-<strong>Input:</strong> <code>38</code>
-<strong>Output:</strong> 2 
-<strong>Explanation: </strong>The process is like: <code>3 + 8 = 11</code>, <code>1 + 1 = 2</code>. 
-&nbsp;            Since <code>2</code> has only one digit, return it.
+<strong>Input:</strong> num = 38
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The process is
+38 --&gt; 3 + 8 --&gt; 11
+11 --&gt; 1 + 1 --&gt; 2 
+Since 2 has only one digit, return it.
 </pre>
 
-<p><b>Follow up:</b><br />
-Could you do it without any loop/recursion in O(1) runtime?</p>
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> num = 0
+<strong>Output:</strong> 0
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>0 &lt;= num &lt;= 2<sup>31</sup> - 1</code></li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong> Could you do it without any loop/recursion in <code>O(1)</code> runtime?</p>
 
 ### Related Topics
   [[Math](../../tag/math/README.md)]

Diff for: problems/add-digits/README.md

@@ -11,68 +11,44 @@
 
 ## [623. Add One Row to Tree (Medium)](https://leetcode.com/problems/add-one-row-to-tree "在二叉树中增加一行")
 
-<p>Given the root of a binary tree, then value <code>v</code> and depth <code>d</code>, you need to add a row of nodes with value <code>v</code> at the given depth <code>d</code>. The root node is at depth 1. </p>
+<p>Given the <code>root</code> of a binary tree and two integers <code>val</code> and <code>depth</code>, add a row of nodes with value <code>val</code> at the given depth <code>depth</code>.</p>
 
-<p>The adding rule is: given a positive integer depth <code>d</code>, for each NOT null tree nodes <code>N</code> in depth <code>d-1</code>, create two tree nodes with value <code>v</code> as <code>N's</code> left subtree root and right subtree root. And <code>N's</code> <b>original left subtree</b> should be the left subtree of the new left subtree root, its <b>original right subtree</b> should be the right subtree of the new right subtree root. If depth <code>d</code> is 1 that means there is no depth d-1 at all, then create a tree node with value <b>v</b> as the new root of the whole original tree, and the original tree is the new root's left subtree.</p>
+<p>Note that the <code>root</code> node is at depth <code>1</code>.</p>
 
-<p><b>Example 1:</b><br />
-<pre>
-<b>Input:</b> 
-A binary tree as following:
-       4
-     /   \
-    2     6
-   / \   / 
-  3   1 5   
-
-<b>v = 1</b>
-
-<b>d = 2</b>
+<p>The adding rule is:</p>
 
-<b>Output:</b> 
-       4
-      / \
-     1   1
-    /     \
-   2       6
-  / \     / 
- 3   1   5   
+<ul>
+	<li>Given the integer <code>depth</code>, for each not null tree node <code>cur</code> at the depth <code>depth - 1</code>, create two tree nodes with value <code>val</code> as <code>cur</code>&#39;s left subtree root and right subtree root.</li>
+	<li><code>cur</code>&#39;s original left subtree should be the left subtree of the new left subtree root.</li>
+	<li><code>cur</code>&#39;s original right subtree should be the right subtree of the new right subtree root.</li>
+	<li>If <code>depth == 1</code> that means there is no depth <code>depth - 1</code> at all, then create a tree node with value <code>val</code> as the new root of the whole original tree, and the original tree is the new root&#39;s left subtree.</li>
+</ul>
 
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/15/addrow-tree.jpg" style="width: 500px; height: 231px;" />
+<pre>
+<strong>Input:</strong> root = [4,2,6,3,1,5], val = 1, depth = 2
+<strong>Output:</strong> [4,1,1,2,null,null,6,3,1,5]
 </pre>
-</p>
-
 
-<p><b>Example 2:</b><br />
+<p><strong>Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/11/add2-tree.jpg" style="width: 500px; height: 277px;" />
 <pre>
-<b>Input:</b> 
-A binary tree as following:
-      4
-     /   
-    2    
-   / \   
-  3   1    
-
-<b>v = 1</b>
-
-<b>d = 3</b>
-
-<b>Output:</b> 
-      4
-     /   
-    2
-   / \    
-  1   1
- /     \  
-3       1
+<strong>Input:</strong> root = [4,2,null,3,1], val = 1, depth = 3
+<strong>Output:</strong> [4,2,null,1,1,3,null,null,1]
 </pre>
-</p>
 
-<p><b>Note:</b><br>
-<ol>
-<li>The given d is in range [1, maximum depth of the given tree + 1].</li>
-<li>The given binary tree has at least one tree node.</li>
-</ol>
-</p>
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
+	<li>The depth of the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
+	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
+	<li><code>-10<sup>5</sup> &lt;= val &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= depth &lt;= the depth of tree + 1</code></li>
+</ul>
 
 ### Related Topics
   [[Tree](../../tag/tree/README.md)]

Diff for: problems/add-one-row-to-tree/README.md

@@ -42,8 +42,9 @@ The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt; prices.length &lt;= 5 * 10<sup>4</sup></code></li>
-	<li><code>0 &lt; prices[i], fee &lt; 5 * 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= prices.length &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= prices[i] &lt; 5 * 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= fee &lt; 5 * 10<sup>4</sup></code></li>
 </ul>
 
 ### Related Topics

Diff for: problems/best-time-to-buy-and-sell-stock-with-transaction-fee/README.md

@@ -11,26 +11,33 @@
 
 ## [257. Binary Tree Paths (Easy)](https://leetcode.com/problems/binary-tree-paths "二叉树的所有路径")
 
-<p>Given a binary tree, return all root-to-leaf paths.</p>
+<p>Given the <code>root</code> of a binary tree, return <em>all root-to-leaf paths in <strong>any order</strong></em>.</p>
 
-<p><strong>Note:</strong>&nbsp;A leaf is a node with no children.</p>
-
-<p><strong>Example:</strong></p>
+<p>A <strong>leaf</strong> is a node with no children.</p>
 
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/12/paths-tree.jpg" style="width: 207px; height: 293px;" />
 <pre>
-<strong>Input:</strong>
-
-   1
- /   \
-2     3
- \
-  5
+<strong>Input:</strong> root = [1,2,3,null,5]
+<strong>Output:</strong> [&quot;1-&gt;2-&gt;5&quot;,&quot;1-&gt;3&quot;]
+</pre>
 
-<strong>Output:</strong> [&quot;1-&gt;2-&gt;5&quot;, &quot;1-&gt;3&quot;]
+<p><strong>Example 2:</strong></p>
 
-<strong>Explanation:</strong> All root-to-leaf paths are: 1-&gt;2-&gt;5, 1-&gt;3
+<pre>
+<strong>Input:</strong> root = [1]
+<strong>Output:</strong> [&quot;1&quot;]
 </pre>
 
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[1, 100]</code>.</li>
+	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
+</ul>
+
 ### Related Topics
   [[Tree](../../tag/tree/README.md)]
   [[Depth-first Search](../../tag/depth-first-search/README.md)]

Diff for: problems/binary-tree-paths/README.md

@@ -38,4 +38,5 @@
 <ul>
 	<li><code>1 &lt;= arr.length &lt;= 1000</code></li>
 	<li><code>2 &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
+	<li>All the values of <code>arr</code> are <strong>unique</strong>.</li>
 </ul>

Diff for: problems/binary-trees-with-factors/README.md

@@ -11,40 +11,49 @@
 
 ## [1252. Cells with Odd Values in a Matrix (Easy)](https://leetcode.com/problems/cells-with-odd-values-in-a-matrix "奇数值单元格的数目")
 
-<p>Given&nbsp;<code>n</code>&nbsp;and&nbsp;<code>m</code>&nbsp;which are the dimensions of a matrix initialized by zeros and given an array <code>indices</code>&nbsp;where <code>indices[i] = [ri, ci]</code>. For each pair of <code>[ri, ci]</code>&nbsp;you have to increment all cells in row <code>ri</code> and column <code>ci</code>&nbsp;by 1.</p>
+<p>There is an <code>m x n</code> matrix that is initialized to all <code>0</code>&#39;s. There is also a 2D array <code>indices</code> where each <code>indices[i] = [r<sub>i</sub>, c<sub>i</sub>]</code> represents a <strong>0-indexed location</strong> to perform some increment operations on the matrix.</p>
 
-<p>Return <em>the number of cells with odd values</em> in the matrix after applying the increment to all <code>indices</code>.</p>
+<p>For each location <code>indices[i]</code>, do <strong>both</strong> of the following:</p>
+
+<ol>
+	<li>Increment <strong>all</strong> the cells on row <code>r<sub>i</sub></code>.</li>
+	<li>Increment <strong>all</strong> the cells on column <code>c<sub>i</sub></code>.</li>
+</ol>
+
+<p>Given <code>m</code>, <code>n</code>, and <code>indices</code>, return <em>the <strong>number of odd-valued cells</strong> in the matrix after applying the increment to all locations in </em><code>indices</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 <img alt="" src="https://assets.leetcode.com/uploads/2019/10/30/e1.png" style="width: 600px; height: 118px;" />
 <pre>
-<strong>Input:</strong> n = 2, m = 3, indices = [[0,1],[1,1]]
+<strong>Input:</strong> m = 2, n = 3, indices = [[0,1],[1,1]]
 <strong>Output:</strong> 6
 <strong>Explanation:</strong> Initial matrix = [[0,0,0],[0,0,0]].
 After applying first increment it becomes [[1,2,1],[0,1,0]].
-The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.
+The final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers.
 </pre>
 
 <p><strong>Example 2:</strong></p>
 <img alt="" src="https://assets.leetcode.com/uploads/2019/10/30/e2.png" style="width: 600px; height: 150px;" />
 <pre>
-<strong>Input:</strong> n = 2, m = 2, indices = [[1,1],[0,0]]
+<strong>Input:</strong> m = 2, n = 2, indices = [[1,1],[0,0]]
 <strong>Output:</strong> 0
-<strong>Explanation:</strong> Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.
+<strong>Explanation:</strong> Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix.
 </pre>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= n &lt;= 50</code></li>
-	<li><code>1 &lt;= m &lt;= 50</code></li>
+	<li><code>1 &lt;= m, n &lt;= 50</code></li>
 	<li><code>1 &lt;= indices.length &lt;= 100</code></li>
-	<li><code>0 &lt;= indices[i][0] &lt;&nbsp;n</code></li>
-	<li><code>0 &lt;= indices[i][1] &lt;&nbsp;m</code></li>
+	<li><code>0 &lt;= r<sub>i</sub> &lt; m</code></li>
+	<li><code>0 &lt;= c<sub>i</sub> &lt; n</code></li>
 </ul>
 
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong> Could you solve this in <code>O(n + m + indices.length)</code> time with only <code>O(n + m)</code> extra space?</p>
+
 ### Related Topics
   [[Array](../../tag/array/README.md)]
 

Diff for: problems/cells-with-odd-values-in-a-matrix/README.md

@@ -13,7 +13,7 @@
 
 <p>Given a binary string <code>s</code> and an integer <code>k</code>.</p>
 
-<p>Return <em>True</em> if every&nbsp;binary code&nbsp;of length <code>k</code> is a substring of <code>s</code>. Otherwise, return <em>False</em>.</p>
+<p>Return <code>true</code> <em>if every binary code of length</em> <code>k</code> <em>is a substring of</em> <code>s</code>. Otherwise, return <code>false</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
@@ -58,8 +58,8 @@
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= s.length &lt;= 5 * 10^5</code></li>
-	<li><code>s</code> consists of 0&#39;s and 1&#39;s only.</li>
+	<li><code>1 &lt;= s.length &lt;= 5 * 10<sup>5</sup></code></li>
+	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
 	<li><code>1 &lt;= k &lt;= 20</code></li>
 </ul>