update

Author: coder-xiaomo

README.md

@@ -1,6 +1,6 @@
 # 力扣题库（完整版）
 
-> 最后更新日期： **2022.05.13**
+> 最后更新日期： **2022.05.24**
 >
 > 使用脚本前请务必仔细完整阅读本 `README.md` 文件
 

leetcode-cn/origin-data.json

@@ -0,0 +1,30 @@
+<p>给你一个字符串 <code>s</code> 和一个字符 <code>letter</code> ，返回在 <code>s</code> 中等于&nbsp;<code>letter</code>&nbsp;字符所占的 <strong>百分比</strong> ，向下取整到最接近的百分比。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "foobar", letter = "o"
+<strong>输出：</strong>33
+<strong>解释：</strong>
+等于字母 'o' 的字符在 s 中占到的百分比是 2 / 6 * 100% = 33% ，向下取整，所以返回 33 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "jjjj", letter = "k"
+<strong>输出：</strong>0
+<strong>解释：</strong>
+等于字母 'k' 的字符在 s 中占到的百分比是 0% ，所以返回 0 。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> 由小写英文字母组成</li>
+	<li><code>letter</code> 是一个小写英文字母</li>
+</ul>

leetcode-cn/originData/maximum-bags-with-full-capacity-of-rocks.json

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+<p>作为国王的统治者，你有一支巫师军队听你指挥。</p>
+
+<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>strength</code>&nbsp;，其中&nbsp;<code>strength[i]</code>&nbsp;表示第&nbsp;<code>i</code>&nbsp;位巫师的力量值。对于连续的一组巫师（也就是这些巫师的力量值是&nbsp;<code>strength</code>&nbsp;的&nbsp;<strong>子数组</strong>），<strong>总力量</strong>&nbsp;定义为以下两个值的&nbsp;<strong>乘积</strong>&nbsp;：</p>
+
+<ul>
+	<li>巫师中 <strong>最弱</strong>&nbsp;的能力值。</li>
+	<li>组中所有巫师的个人力量值 <strong>之和</strong>&nbsp;。</li>
+</ul>
+
+<p>请你返回 <strong>所有</strong>&nbsp;巫师组的 <strong>总</strong>&nbsp;力量之和。由于答案可能很大，请将答案对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;<strong>取余</strong>&nbsp;后返回。</p>
+
+<p><strong>子数组</strong>&nbsp;是一个数组里 <strong>非空</strong>&nbsp;连续子序列。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>strength = [1,3,1,2]
+<b>输出：</b>44
+<b>解释：</b>以下是所有连续巫师组：
+- [<em><strong>1</strong></em>,3,1,2] 中 [1] ，总力量值为 min([1]) * sum([1]) = 1 * 1 = 1
+- [1,<em><strong>3</strong></em>,1,2] 中 [3] ，总力量值为 min([3]) * sum([3]) = 3 * 3 = 9
+- [1,3,<em><strong>1</strong></em>,2] 中 [1] ，总力量值为 min([1]) * sum([1]) = 1 * 1 = 1
+- [1,3,1,<em><strong>2</strong></em>] 中 [2] ，总力量值为 min([2]) * sum([2]) = 2 * 2 = 4
+- [<em><strong>1,3</strong></em>,1,2] 中 [1,3] ，总力量值为 min([1,3]) * sum([1,3]) = 1 * 4 = 4
+- [1,<em><strong>3,1</strong></em>,2] 中 [3,1] ，总力量值为 min([3,1]) * sum([3,1]) = 1 * 4 = 4
+- [1,3,<em><strong>1,2</strong></em>] 中 [1,2] ，总力量值为 min([1,2]) * sum([1,2]) = 1 * 3 = 3
+- [<em><strong>1,3,1</strong></em>,2] 中 [1,3,1] ，总力量值为 min([1,3,1]) * sum([1,3,1]) = 1 * 5 = 5
+- [1,<em><strong>3,1,2</strong></em>] 中 [3,1,2] ，总力量值为 min([3,1,2]) * sum([3,1,2]) = 1 * 6 = 6
+- [<em><strong>1,3,1,2</strong></em>] 中 [1,3,1,2] ，总力量值为 min([1,3,1,2]) * sum([1,3,1,2]) = 1 * 7 = 7
+所有力量值之和为 1 + 9 + 1 + 4 + 4 + 4 + 3 + 5 + 6 + 7 = 44 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>strength = [5,4,6]
+<b>输出：</b>213
+<b>解释：</b>以下是所有连续巫师组：
+- [<em><strong>5</strong></em>,4,6] 中 [5] ，总力量值为 min([5]) * sum([5]) = 5 * 5 = 25
+- [5,<em><strong>4</strong></em>,6] 中 [4] ，总力量值为 min([4]) * sum([4]) = 4 * 4 = 16
+- [5,4,<em><strong>6</strong></em>] 中 [6] ，总力量值为 min([6]) * sum([6]) = 6 * 6 = 36
+- [<em><strong>5,4</strong></em>,6] 中 [5,4] ，总力量值为 min([5,4]) * sum([5,4]) = 4 * 9 = 36
+- [5,<em><strong>4,6</strong></em>] 中 [4,6] ，总力量值为 min([4,6]) * sum([4,6]) = 4 * 10 = 40
+- [<em><strong>5,4,6</strong></em>] 中 [5,4,6] ，总力量值为 min([5,4,6]) * sum([5,4,6]) = 4 * 15 = 60
+所有力量值之和为 25 + 16 + 36 + 36 + 40 + 60 = 213 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= strength.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= strength[i] &lt;= 10<sup>9</sup></code></li>
+</ul>

leetcode-cn/originData/minimum-lines-to-represent-a-line-chart.json

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+<p>给你一个二维整数数组&nbsp;<code>stockPrices</code> ，其中&nbsp;<code>stockPrices[i] = [day<sub>i</sub>, price<sub>i</sub>]</code>&nbsp;表示股票在&nbsp;<code>day<sub>i</sub></code>&nbsp;的价格为&nbsp;<code>price<sub>i</sub></code>&nbsp;。<strong>折线图</strong>&nbsp;是一个二维平面上的若干个点组成的图，横坐标表示日期，纵坐标表示价格，折线图由相邻的点连接而成。比方说下图是一个例子：</p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/03/30/1920px-pushkin_population_historysvg.png" style="width: 500px; height: 313px;">
+<p>请你返回要表示一个折线图所需要的 <strong>最少线段数</strong>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2022/03/30/ex0.png" style="width: 400px; height: 400px;"></p>
+
+<pre><b>输入：</b>stockPrices = [[1,7],[2,6],[3,5],[4,4],[5,4],[6,3],[7,2],[8,1]]
+<b>输出：</b>3
+<strong>解释：</strong>
+上图为输入对应的图，横坐标表示日期，纵坐标表示价格。
+以下 3 个线段可以表示折线图：
+- 线段 1 （红色）从 (1,7) 到 (4,4) ，经过 (1,7) ，(2,6) ，(3,5) 和 (4,4) 。
+- 线段 2 （蓝色）从 (4,4) 到 (5,4) 。
+- 线段 3 （绿色）从 (5,4) 到 (8,1) ，经过 (5,4) ，(6,3) ，(7,2) 和 (8,1) 。
+可以证明，无法用少于 3 条线段表示这个折线图。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2022/03/30/ex1.png" style="width: 325px; height: 325px;"></p>
+
+<pre><b>输入：</b>stockPrices = [[3,4],[1,2],[7,8],[2,3]]
+<b>输出：</b>1
+<strong>解释：</strong>
+如上图所示，折线图可以用一条线段表示。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= stockPrices.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>stockPrices[i].length == 2</code></li>
+	<li><code>1 &lt;= day<sub>i</sub>, price<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
+	<li>所有&nbsp;<code>day<sub>i</sub></code>&nbsp;<strong>互不相同</strong>&nbsp;。</li>
+</ul>

leetcode-cn/originData/percentage-of-letter-in-string.json

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+<p>现有编号从&nbsp;<code>0</code> 到 <code>n - 1</code> 的 <code>n</code> 个背包。给你两个下标从 <strong>0</strong> 开始的整数数组 <code>capacity</code> 和 <code>rocks</code> 。第 <code>i</code> 个背包最大可以装 <code>capacity[i]</code> 块石头，当前已经装了 <code>rocks[i]</code> 块石头。另给你一个整数 <code>additionalRocks</code> ，表示<span class="text-only" data-eleid="10" style="white-space: pre;">你可以放置的额外石头数量，石头可以往 </span><strong><span class="text-only" data-eleid="11" style="white-space: pre;">任意</span></strong><span class="text-only" data-eleid="12" style="white-space: pre;"> 背包中放置。</span></p>
+
+<p>请你将额外的石头放入一些背包中，并返回放置后装满石头的背包的 <strong>最大 </strong>数量<em>。</em></p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>capacity = [2,3,4,5], rocks = [1,2,4,4], additionalRocks = 2
+<strong>输出：</strong>3
+<strong>解释：</strong>
+1 块石头放入背包 0 ，1 块石头放入背包 1 。
+每个背包中的石头总数是 [2,3,4,4] 。
+背包 0 、背包 1 和 背包 2 都装满石头。
+总计 3 个背包装满石头，所以返回 3 。
+可以证明不存在超过 3 个背包装满石头的情况。
+注意，可能存在其他放置石头的方案同样能够得到 3 这个结果。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>capacity = [10,2,2], rocks = [2,2,0], additionalRocks = 100
+<strong>输出：</strong>3
+<strong>解释：</strong>
+8 块石头放入背包 0 ，2 块石头放入背包 2 。
+每个背包中的石头总数是 [10,2,2] 。
+背包 0 、背包 1 和背包 2 都装满石头。
+总计 3 个背包装满石头，所以返回 3 。
+可以证明不存在超过 3 个背包装满石头的情况。
+注意，不必用完所有的额外石头。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>n == capacity.length == rocks.length</code></li>
+	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= capacity[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= rocks[i] &lt;= capacity[i]</code></li>
+	<li><code>1 &lt;= additionalRocks &lt;= 10<sup>9</sup></code></li>
+</ul>

leetcode-cn/originData/sum-of-total-strength-of-wizards.json

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+<p>Given a string <code>s</code> and a character <code>letter</code>, return<em> the <strong>percentage</strong> of characters in </em><code>s</code><em> that equal </em><code>letter</code><em> <strong>rounded down</strong> to the nearest whole percent.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;foobar&quot;, letter = &quot;o&quot;
+<strong>Output:</strong> 33
+<strong>Explanation:</strong>
+The percentage of characters in s that equal the letter &#39;o&#39; is 2 / 6 * 100% = 33% when rounded down, so we return 33.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;jjjj&quot;, letter = &quot;k&quot;
+<strong>Output:</strong> 0
+<strong>Explanation:</strong>
+The percentage of characters in s that equal the letter &#39;k&#39; is 0%, so we return 0.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> consists of lowercase English letters.</li>
+	<li><code>letter</code> is a lowercase English letter.</li>
+</ul>

leetcode-cn/problem (Chinese)/字母在字符串中的百分比 [percentage-of-letter-in-string].html

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+<p>As the ruler of a kingdom, you have an army of wizards at your command.</p>
+
+<p>You are given a <strong>0-indexed</strong> integer array <code>strength</code>, where <code>strength[i]</code> denotes the strength of the <code>i<sup>th</sup></code> wizard. For a <strong>contiguous</strong> group of wizards (i.e. the wizards&#39; strengths form a <strong>subarray</strong> of <code>strength</code>), the <strong>total strength</strong> is defined as the <strong>product</strong> of the following two values:</p>
+
+<ul>
+	<li>The strength of the <strong>weakest</strong> wizard in the group.</li>
+	<li>The <strong>total</strong> of all the individual strengths of the wizards in the group.</li>
+</ul>
+
+<p>Return <em>the <strong>sum</strong> of the total strengths of <strong>all</strong> contiguous groups of wizards</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>A <strong>subarray</strong> is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> strength = [1,3,1,2]
+<strong>Output:</strong> 44
+<strong>Explanation:</strong> The following are all the contiguous groups of wizards:
+- [1] from [<u><strong>1</strong></u>,3,1,2] has a total strength of min([1]) * sum([1]) = 1 * 1 = 1
+- [3] from [1,<u><strong>3</strong></u>,1,2] has a total strength of min([3]) * sum([3]) = 3 * 3 = 9
+- [1] from [1,3,<u><strong>1</strong></u>,2] has a total strength of min([1]) * sum([1]) = 1 * 1 = 1
+- [2] from [1,3,1,<u><strong>2</strong></u>] has a total strength of min([2]) * sum([2]) = 2 * 2 = 4
+- [1,3] from [<u><strong>1,3</strong></u>,1,2] has a total strength of min([1,3]) * sum([1,3]) = 1 * 4 = 4
+- [3,1] from [1,<u><strong>3,1</strong></u>,2] has a total strength of min([3,1]) * sum([3,1]) = 1 * 4 = 4
+- [1,2] from [1,3,<u><strong>1,2</strong></u>] has a total strength of min([1,2]) * sum([1,2]) = 1 * 3 = 3
+- [1,3,1] from [<u><strong>1,3,1</strong></u>,2] has a total strength of min([1,3,1]) * sum([1,3,1]) = 1 * 5 = 5
+- [3,1,2] from [1,<u><strong>3,1,2</strong></u>] has a total strength of min([3,1,2]) * sum([3,1,2]) = 1 * 6 = 6
+- [1,3,1,2] from [<u><strong>1,3,1,2</strong></u>] has a total strength of min([1,3,1,2]) * sum([1,3,1,2]) = 1 * 7 = 7
+The sum of all the total strengths is 1 + 9 + 1 + 4 + 4 + 4 + 3 + 5 + 6 + 7 = 44.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> strength = [5,4,6]
+<strong>Output:</strong> 213
+<strong>Explanation:</strong> The following are all the contiguous groups of wizards: 
+- [5] from [<u><strong>5</strong></u>,4,6] has a total strength of min([5]) * sum([5]) = 5 * 5 = 25
+- [4] from [5,<u><strong>4</strong></u>,6] has a total strength of min([4]) * sum([4]) = 4 * 4 = 16
+- [6] from [5,4,<u><strong>6</strong></u>] has a total strength of min([6]) * sum([6]) = 6 * 6 = 36
+- [5,4] from [<u><strong>5,4</strong></u>,6] has a total strength of min([5,4]) * sum([5,4]) = 4 * 9 = 36
+- [4,6] from [5,<u><strong>4,6</strong></u>] has a total strength of min([4,6]) * sum([4,6]) = 4 * 10 = 40
+- [5,4,6] from [<u><strong>5,4,6</strong></u>] has a total strength of min([5,4,6]) * sum([5,4,6]) = 4 * 15 = 60
+The sum of all the total strengths is 25 + 16 + 36 + 36 + 40 + 60 = 213.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= strength.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= strength[i] &lt;= 10<sup>9</sup></code></li>
+</ul>

leetcode-cn/problem (Chinese)/巫师的总力量和 [sum-of-total-strength-of-wizards].html

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+<p>You are given a 2D integer array <code>stockPrices</code> where <code>stockPrices[i] = [day<sub>i</sub>, price<sub>i</sub>]</code> indicates the price of the stock on day <code>day<sub>i</sub></code> is <code>price<sub>i</sub></code>. A <strong>line chart</strong> is created from the array by plotting the points on an XY plane with the X-axis representing the day and the Y-axis representing the price and connecting adjacent points. One such example is shown below:</p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/03/30/1920px-pushkin_population_historysvg.png" style="width: 500px; height: 313px;" />
+<p>Return <em>the <strong>minimum number of lines</strong> needed to represent the line chart</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/03/30/ex0.png" style="width: 400px; height: 400px;" />
+<pre>
+<strong>Input:</strong> stockPrices = [[1,7],[2,6],[3,5],[4,4],[5,4],[6,3],[7,2],[8,1]]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong>
+The diagram above represents the input, with the X-axis representing the day and Y-axis representing the price.
+The following 3 lines can be drawn to represent the line chart:
+- Line 1 (in red) from (1,7) to (4,4) passing through (1,7), (2,6), (3,5), and (4,4).
+- Line 2 (in blue) from (4,4) to (5,4).
+- Line 3 (in green) from (5,4) to (8,1) passing through (5,4), (6,3), (7,2), and (8,1).
+It can be shown that it is not possible to represent the line chart using less than 3 lines.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/03/30/ex1.png" style="width: 325px; height: 325px;" />
+<pre>
+<strong>Input:</strong> stockPrices = [[3,4],[1,2],[7,8],[2,3]]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong>
+As shown in the diagram above, the line chart can be represented with a single line.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= stockPrices.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>stockPrices[i].length == 2</code></li>
+	<li><code>1 &lt;= day<sub>i</sub>, price<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
+	<li>All <code>day<sub>i</sub></code> are <strong>distinct</strong>.</li>
+</ul>

leetcode-cn/problem (Chinese)/表示一个折线图的最少线段数 [minimum-lines-to-represent-a-line-chart].html

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+<p>You have <code>n</code> bags numbered from <code>0</code> to <code>n - 1</code>. You are given two <strong>0-indexed</strong> integer arrays <code>capacity</code> and <code>rocks</code>. The <code>i<sup>th</sup></code> bag can hold a maximum of <code>capacity[i]</code> rocks and currently contains <code>rocks[i]</code> rocks. You are also given an integer <code>additionalRocks</code>, the number of additional rocks you can place in <strong>any</strong> of the bags.</p>
+
+<p>Return<em> the <strong>maximum</strong> number of bags that could have full capacity after placing the additional rocks in some bags.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> capacity = [2,3,4,5], rocks = [1,2,4,4], additionalRocks = 2
+<strong>Output:</strong> 3
+<strong>Explanation:</strong>
+Place 1 rock in bag 0 and 1 rock in bag 1.
+The number of rocks in each bag are now [2,3,4,4].
+Bags 0, 1, and 2 have full capacity.
+There are 3 bags at full capacity, so we return 3.
+It can be shown that it is not possible to have more than 3 bags at full capacity.
+Note that there may be other ways of placing the rocks that result in an answer of 3.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> capacity = [10,2,2], rocks = [2,2,0], additionalRocks = 100
+<strong>Output:</strong> 3
+<strong>Explanation:</strong>
+Place 8 rocks in bag 0 and 2 rocks in bag 2.
+The number of rocks in each bag are now [10,2,2].
+Bags 0, 1, and 2 have full capacity.
+There are 3 bags at full capacity, so we return 3.
+It can be shown that it is not possible to have more than 3 bags at full capacity.
+Note that we did not use all of the additional rocks.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == capacity.length == rocks.length</code></li>
+	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= capacity[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= rocks[i] &lt;= capacity[i]</code></li>
+	<li><code>1 &lt;= additionalRocks &lt;= 10<sup>9</sup></code></li>
+</ul>