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‎solution/0000-0099/0064.Minimum Path Sum/README.md

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@@ -6,36 +6,36 @@
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<!-- 这里写题目描述 -->
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<p>给定一个包含非负整数的 <code><em>m</em> x <em>n</em></code> 网格 <code>grid</code> ，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。</p>
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<p>给定一个包含非负整数的 <code><em>m</em>&nbsp;x&nbsp;<em>n</em></code>&nbsp;网格&nbsp;<code>grid</code> ，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。</p>
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<p><strong>说明：</strong>每次只能向下或者向右移动一步。</p>
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<p> </p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<p><strong class="example">示例 1：</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0064.Minimum%20Path%20Sum/images/minpath.jpg" style="width: 242px; height: 242px;" />
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<pre>
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<strong>输入：</strong>grid = [[1,3,1],[1,5,1],[4,2,1]]
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<strong>输出：</strong>7
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<strong>解释：</strong>因为路径 1→3→1→1→1 的总和最小。
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</pre>
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<p><strong>示例 2：</strong></p>
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<p><strong class="example">示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>grid = [[1,2,3],[4,5,6]]
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<strong>输出：</strong>12
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</pre>
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<p> </p>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>m == grid.length</code></li>
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<li><code>n == grid[i].length</code></li>
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<li><code>1 <= m, n <= 200</code></li>
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<li><code>0 <= grid[i][j] <= 100</code></li>
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<li><code>1 &lt;= m, n &lt;= 200</code></li>
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<li><code>0 &lt;= grid[i][j] &lt;= 200</code></li>
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</ul>
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## 解法

‎solution/0000-0099/0064.Minimum Path Sum/README_EN.md

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<li><code>m == grid.length</code></li>
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<li><code>n == grid[i].length</code></li>
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<li><code>1 &lt;= m, n &lt;= 200</code></li>
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<li><code>0 &lt;= grid[i][j] &lt;= 100</code></li>
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<li><code>0 &lt;= grid[i][j] &lt;= 200</code></li>
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</ul>
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## Solutions

‎solution/0100-0199/0195.Tenth Line/README_EN.md

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<p>Assume that <code>file.txt</code> has the following content:</p>
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<pre>
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Line 1
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Line 2
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Line 3
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Line 4
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Line 5
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</pre>
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<p>Your script should output the tenth line, which is:</p>
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<pre>
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Line 10
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</pre>
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<div class="spoilers"><b>Note:</b><br />
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1. If the file contains less than 10 lines, what should you output?<br />
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2. There&#39;s at least three different solutions. Try to explore all possibilities.</div>
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## Solutions

‎solution/0200-0299/0274.H-Index/README.md

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<p>给你一个整数数组 <code>citations</code> ，其中 <code>citations[i]</code> 表示研究者的第 <code>i</code> 篇论文被引用的次数。计算并返回该研究者的 <strong><code>h</code><em>&nbsp;</em>指数</strong>。</p>
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<p>根据维基百科上&nbsp;<a href="https://baike.baidu.com/item/h-index/3991452?fr=aladdin" target="_blank">h 指数的定义</a>：h 代表“高引用次数”，一名科研人员的 <code>h</code><strong>指数</strong>是指他（她）的 （<code>n</code> 篇论文中）<strong>总共</strong>有 <code>h</code> 篇论文分别被引用了<strong>至少</strong> <code>h</code> 次。且其余的 <em><code>n - h</code>&nbsp;</em>篇论文每篇被引用次数&nbsp;<strong>不超过 </strong><em><code>h</code> </em>次。</p>
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<p>如果 <code>h</code><em> </em>有多种可能的值，<strong><code>h</code> 指数 </strong>是其中最大的那个。</p>
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<p>根据维基百科上&nbsp;<a href="https://baike.baidu.com/item/h-index/3991452?fr=aladdin" target="_blank">h 指数的定义</a>：<code>h</code> 代表“高引用次数” ，一名科研人员的 <code>h</code><strong> 指数 </strong>是指他（她）至少发表了 <code>h</code> 篇论文，并且每篇论文<strong> 至少</strong> 被引用 <code>h</code> 次。如果 <code>h</code><em> </em>有多种可能的值，<strong><code>h</code> 指数 </strong>是其中最大的那个。</p>
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<p>&nbsp;</p>
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‎solution/0300-0399/0327.Count of Range Sum/Solution.cpp

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Original file line numberDiff line numberDiff line change
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class BinaryIndexedTree {
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public:
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BinaryIndexedTree(int _n) : n(_n), c(_n + 1) {}
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BinaryIndexedTree(int _n)
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: n(_n)
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, c(_n + 1) {}
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void update(int x, int v) {
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while (x <= n) {

‎solution/0300-0399/0328.Odd Even Linked List/Solution.cpp

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if (!head) {
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return nullptr;
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}
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ListNode *a = head;
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ListNode* a = head;
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ListNode *b = head->next, *c = b;
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while (b && b->next) {
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a->next = b->next;

‎solution/0300-0399/0329.Longest Increasing Path in a Matrix/README.md

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this.matrix = matrix;
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int ans = 0;
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for (int i = 0; i < m; ++i) {
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for (int j = 0;j < n; ++j) {
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for (int j = 0; j < n; ++j) {
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ans = Math.max(ans, dfs(i, j));
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}
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}

‎solution/0300-0399/0329.Longest Increasing Path in a Matrix/README_EN.md

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this.matrix = matrix;
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int ans = 0;
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for (int i = 0; i < m; ++i) {
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for (int j = 0;j < n; ++j) {
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for (int j = 0; j < n; ++j) {
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ans = Math.max(ans, dfs(i, j));
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}
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}

‎solution/0300-0399/0329.Longest Increasing Path in a Matrix/Solution.java

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this.matrix = matrix;
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int ans = 0;
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for (int i = 0; i < m; ++i) {
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for (int j = 0;j < n; ++j) {
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for (int j = 0; j < n; ++j) {
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ans = Math.max(ans, dfs(i, j));
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}
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}

‎solution/0300-0399/0336.Palindrome Pairs/README.md

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<strong>输出：</strong>[[0,1],[1,0]]
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</pre>
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<p><strong>提示：</strong></p>
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<ul>

‎solution/0500-0599/0543.Diameter of Binary Tree/README.md

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<!-- 这里写题目描述 -->
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<p>给定一棵二叉树，你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过也可能不穿过根结点。</p>
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<p>给你一棵二叉树的根节点，返回该树的 <strong>直径</strong> 。</p>
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<p>&nbsp;</p>
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<p>二叉树的 <strong>直径</strong> 是指树中任意两个节点之间最长路径的 <strong>长度</strong> 。这条路径可能经过也可能不经过根节点 <code>root</code> 。</p>
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<p>两节点之间路径的 <strong>长度</strong> 由它们之间边数表示。</p>
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<p><strong>示例 :</strong><br>
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给定二叉树</p>
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<p>&nbsp;</p>
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<pre> 1
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/ \
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2 3
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/ \
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4 5
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<p><strong class="example">示例 1：</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0543.Diameter%20of%20Binary%20Tree/images/diamtree.jpg" style="width: 292px; height: 302px;" />
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<pre>
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<strong>输入：</strong>root = [1,2,3,4,5]
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<strong>输出：</strong>3
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<strong>解释：</strong>3 ，取路径 [4,2,1,3] 或 [5,2,1,3] 的长度。
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</pre>
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<p>返回&nbsp;<strong>3</strong>, 它的长度是路径 [4,2,1,3] 或者&nbsp;[5,2,1,3]。</p>
25+
<p><strong class="example">示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>root = [1,2]
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<strong>输出：</strong>1
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</pre>
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<p>&nbsp;</p>
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<p><strong>注意：</strong>两结点之间的路径长度是以它们之间边的数目表示。</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li>树中节点数目在范围 <code>[1, 10<sup>4</sup>]</code> 内</li>
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<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
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</ul>
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## 解法
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‎solution/0700-0799/0756.Pyramid Transition Matrix/README.md

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<p><img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0756.Pyramid%20Transition%20Matrix/images/pyramid2-grid.jpg" style="height: 359px; width: 600px;" /></p>
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<pre>
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<strong>输入：</strong>bottom = "AABA", allowed = ["AAB","AAC","BCD","BBE","DEF"]
40+
<strong>输入：</strong>bottom = "AAAA", allowed = ["AAB","AAC","BCD","BBE","DEF"]
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<strong>输出：</strong>false
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<strong>解释：</strong>允许的三角形模式显示在右边。
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从最底层(游戏邦注:即第4个关卡)开始，创造第3个关卡有多种方法，但如果尝试所有可能性，你便会在创造第1个关卡前陷入困境。

‎solution/0700-0799/0799.Champagne Tower/README_EN.md

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<p>Now after pouring some non-negative integer cups of champagne, return how full the <code>j<sup>th</sup></code> glass in the <code>i<sup>th</sup></code> row is (both <code>i</code> and <code>j</code> are 0-indexed.)</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> poured = 1, query_row = 1, query_glass = 1
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<strong>Output:</strong> 0.00000
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<strong>Explanation:</strong> We poured 1 cup of champange to the top glass of the tower (which is indexed as (0, 0)). There will be no excess liquid so all the glasses under the top glass will remain empty.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> poured = 2, query_row = 1, query_glass = 1
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<strong>Output:</strong> 0.50000
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<strong>Explanation:</strong> We poured 2 cups of champange to the top glass of the tower (which is indexed as (0, 0)). There is one cup of excess liquid. The glass indexed as (1, 0) and the glass indexed as (1, 1) will share the excess liquid equally, and each will get half cup of champange.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> poured = 100000009, query_row = 33, query_glass = 17
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<strong>Output:</strong> 1.00000
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>0 &lt;=&nbsp;poured &lt;= 10<sup>9</sup></code></li>
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<li><code>0 &lt;= query_glass &lt;= query_row&nbsp;&lt; 100</code></li>
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<li><code>0 &lt;=&nbsp;poured &lt;= 10<sup>9</sup></code></li>
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<li><code>0 &lt;= query_glass &lt;= query_row&nbsp;&lt; 100</code></li>
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</ul>
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## Solutions

‎solution/0800-0899/0870.Advantage Shuffle/README.md

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<!-- 这里写题目描述 -->
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<p>给定两个大小相等的数组&nbsp;<code>nums1</code>&nbsp;和&nbsp;<code>nums2</code>，<code>nums1</code>&nbsp;相对于 <code>nums2</code> 的<em>优势</em>可以用满足&nbsp;<code>nums1[i] &gt; nums2[i]</code>&nbsp;的索引 <code>i</code>&nbsp;的数目来描述。</p>
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<p>给定两个长度相等的数组&nbsp;<code>nums1</code>&nbsp;和&nbsp;<code>nums2</code>，<code>nums1</code>&nbsp;相对于 <code>nums2</code> 的<em>优势</em>可以用满足&nbsp;<code>nums1[i] &gt; nums2[i]</code>&nbsp;的索引 <code>i</code>&nbsp;的数目来描述。</p>
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<p>返回 <font color="#c7254e" face="Menlo, Monaco, Consolas, Courier New, monospace" size="1"><span style="background-color: rgb(249, 242, 244);">nums1</span></font>&nbsp;的<strong>任意</strong>排列，使其相对于 <code>nums2</code>&nbsp;的优势最大化。</p>
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‎solution/0900-0999/0930.Binary Subarrays With Sum/README_EN.md

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<p>A <strong>subarray</strong> is a contiguous part of the array.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,0,1,0,1], goal = 2
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> The 4 subarrays are bolded and underlined below:
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[<u><strong>1,0,1</strong></u>,0,1]
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[<u><strong>1,0,1,0</strong></u>,1]
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[1,<u><strong>0,1,0,1</strong></u>]
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[1,0,<u><strong>1,0,1</strong></u>]
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [0,0,0,0,0], goal = 0
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<strong>Output:</strong> 15
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
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<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
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<li><code>0 &lt;= goal &lt;= nums.length</code></li>
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<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
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<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
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<li><code>0 &lt;= goal &lt;= nums.length</code></li>
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</ul>
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## Solutions

‎solution/1000-1099/1080.Insufficient Nodes in Root to Leaf Paths/README.md

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<!-- 这里写题目描述 -->
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<p>给定一棵二叉树的根 <code>root</code>，请你考虑它所有&nbsp;<strong>从根到叶的路径</strong>：从根到任何叶的路径。（所谓一个叶子节点，就是一个没有子节点的节点）</p>
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<p>给你二叉树的根节点 <code>root</code> 和一个整数 <code>limit</code> ，请你同时删除树中所有 <strong>不足节点 </strong>，并返回最终二叉树的根节点。</p>
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<p>假如通过节点 <code>node</code> 的每种可能的 &ldquo;根-叶&rdquo; 路径上值的总和全都小于给定的 <code>limit</code>，则该节点被称之为「不足节点」，需要被删除。</p>
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<p>假如通过节点 <code>node</code> 的每种可能的 “根-叶” 路径上值的总和全都小于给定的 <code>limit</code>，则该节点被称之为<strong> 不足节点 </strong>，需要被删除。</p>
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<p>请你删除所有不足节点，并返回生成的二叉树的根。</p>
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<p><strong>叶子节点</strong>，就是没有子节点的节点。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1080.Insufficient%20Nodes%20in%20Root%20to%20Leaf%20Paths/images/insufficient-1.png" style="height: 200px; width: 482px;">
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输入：</strong>root = [1,2,3,4,-99,-99,7,8,9,-99,-99,12,13,-99,14], limit = 1
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<strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1080.Insufficient%20Nodes%20in%20Root%20to%20Leaf%20Paths/images/insufficient-2.png" style="height: 200px; width: 258px;">
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输出：</strong>[1,2,3,4,null,null,7,8,9,null,14]
17+
<p><strong class="example">示例 1：</strong></p>
18+
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1080.Insufficient%20Nodes%20in%20Root%20to%20Leaf%20Paths/images/insufficient-11.png" style="width: 500px; height: 207px;" />
19+
<pre>
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<strong>输入：</strong>root = [1,2,3,4,-99,-99,7,8,9,-99,-99,12,13,-99,14], limit = 1
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<strong>输出：</strong>[1,2,3,4,null,null,7,8,9,null,14]
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1080.Insufficient%20Nodes%20in%20Root%20to%20Leaf%20Paths/images/insufficient-3.png" style="height: 200px; width: 292px;">
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输入：</strong>root = [5,4,8,11,null,17,4,7,1,null,null,5,3], limit = 22
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<strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1080.Insufficient%20Nodes%20in%20Root%20to%20Leaf%20Paths/images/insufficient-4.png" style="height: 200px; width: 264px;">
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输出：</strong>[5,4,8,11,null,17,4,7,null,null,null,5]</pre>
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<p><strong>示例 3：</strong></p>
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<p><strong class="example">示例 2：</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1080.Insufficient%20Nodes%20in%20Root%20to%20Leaf%20Paths/images/insufficient-3.png" style="width: 400px; height: 274px;" />
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<pre>
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<strong>输入：</strong>root = [5,4,8,11,null,17,4,7,1,null,null,5,3], limit = 22
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<strong>输出：</strong>[5,4,8,11,null,17,4,7,null,null,null,5]
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</pre>
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<pre><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1080.Insufficient%20Nodes%20in%20Root%20to%20Leaf%20Paths/images/insufficient-5.png" style="height: 100px; width: 140px;">
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输入：</strong>root = [5,-6,-6], limit = 0<strong>
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输出：</strong>[]</pre>
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<p><strong class="example">示例 3：</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1080.Insufficient%20Nodes%20in%20Root%20to%20Leaf%20Paths/images/screen-shot-2019-06-11-at-83301-pm.png" style="width: 250px; height: 199px;" />
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<pre>
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<strong>输入：</strong>root = [1,2,-3,-5,null,4,null], limit = -1
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<strong>输出：</strong>[1,null,-3,4]
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ol>
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<li>给定的树有&nbsp;<code>1</code>&nbsp;到&nbsp;<code>5000</code>&nbsp;个节点</li>
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<li><code>-10^5&nbsp;&lt;= node.val &lt;= 10^5</code></li>
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<li><code>-10^9 &lt;= limit&nbsp;&lt;= 10^9</code></li>
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</ol>
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<ul>
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<li>树中节点数目在范围 <code>[1, 5000]</code> 内</li>
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<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
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<li><code>-10<sup>9</sup> &lt;= limit &lt;= 10<sup>9</sup></code></li>
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</ul>
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<p>&nbsp;</p>
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