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LeetCode 题目描述 1554~1569
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‎solution/1500-1599/1554.Strings Differ by One Character/README.md

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# [1554. 只有一个不同字符的字符串](https://leetcode-cn.com/problems/strings-differ-by-one-character)
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[English Version](/solution/1500-1599/1554.Strings Differ by One Character/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给定一个字符串列表&nbsp;<code>dict</code> ，其中所有字符串的长度都相同。</p>
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<p>当存在两个字符串在相同索引处只有一个字符不同时，返回 <code>True</code> ，否则返回 <code>False</code> 。</p>
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<p><strong>进阶：</strong>你可以以 O(n*m) 的复杂度解决问题吗？其中 n 是列表 <code>dict</code> 的长度，m 是字符串的长度。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>dict = [&quot;abcd&quot;,&quot;acbd&quot;, &quot;aacd&quot;]
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<strong>输出：</strong>true
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<strong>解释：</strong>字符串 &quot;a<strong>b</strong>cd&quot; 和 &quot;a<strong>a</strong>cd&quot; 只在索引 1 处有一个不同的字符。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>dict = [&quot;ab&quot;,&quot;cd&quot;,&quot;yz&quot;]
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<strong>输出：</strong>false
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</pre>
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<p><strong>示例 3：</strong></p>
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<pre>
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<strong>输入：</strong>dict = [&quot;abcd&quot;,&quot;cccc&quot;,&quot;abyd&quot;,&quot;abab&quot;]
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<strong>输出：</strong>true
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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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<ul>
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<li><code>dict</code>&nbsp;中的字符数小于或等于&nbsp;<code>10^5</code>&nbsp;。</li>
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<li><code>dict[i].length == dict[j].length</code></li>
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<li><code>dict[i]</code>&nbsp;是互不相同的。</li>
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<li><code>dict[i]</code>&nbsp;只包含小写英文字母。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1500-1599/1554.Strings Differ by One Character/README_EN.md

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# [1554. Strings Differ by One Character](https://leetcode.com/problems/strings-differ-by-one-character)
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[中文文档](/solution/1500-1599/1554.Strings Differ by One Character/README.md)
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## Description
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<p>Given a list&nbsp;of strings <code>dict</code> where all the strings are of the same length.</p>
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<p>Return <code>True</code> if there are 2 strings that only differ by 1 character in the same index, otherwise&nbsp;return <code>False</code>.</p>
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<p><strong>Follow up:&nbsp;</strong>Could you solve this problem in O(n*m) where n is the length of <code>dict</code> and m is the length of each string.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> dict = [&quot;abcd&quot;,&quot;acbd&quot;, &quot;aacd&quot;]
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<strong>Output:</strong> true
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<strong>Explanation:</strong> Strings &quot;a<strong>b</strong>cd&quot; and &quot;a<strong>a</strong>cd&quot; differ only by one character in the index 1.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> dict = [&quot;ab&quot;,&quot;cd&quot;,&quot;yz&quot;]
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<strong>Output:</strong> false
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> dict = [&quot;abcd&quot;,&quot;cccc&quot;,&quot;abyd&quot;,&quot;abab&quot;]
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<strong>Output:</strong> true
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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>Number of characters in <code>dict &lt;= 10^5</code></li>
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<li><code>dict[i].length == dict[j].length</code></li>
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<li><code>dict[i]</code> should be unique.</li>
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<li><code>dict[i]</code> contains only lowercase English letters.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1500-1599/1556.Thousand Separator/README.md

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# [1556. 千位分隔数](https://leetcode-cn.com/problems/thousand-separator)
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[English Version](/solution/1500-1599/1556.Thousand Separator/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个整数&nbsp;<code>n</code>，请你每隔三位添加点（即 &quot;.&quot; 符号）作为千位分隔符，并将结果以字符串格式返回。</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>输入：</strong>n = 987
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<strong>输出：</strong>&quot;987&quot;
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><strong>输入：</strong>n = 1234
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<strong>输出：</strong>&quot;1.234&quot;
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</pre>
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<p><strong>示例 3：</strong></p>
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<pre><strong>输入：</strong>n = 123456789
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<strong>输出：</strong>&quot;123.456.789&quot;
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</pre>
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<p><strong>示例 4：</strong></p>
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<pre><strong>输入：</strong>n = 0
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<strong>输出：</strong>&quot;0&quot;
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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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<ul>
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<li><code>0 &lt;= n &lt; 2^31</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1500-1599/1556.Thousand Separator/README_EN.md

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# [1556. Thousand Separator](https://leetcode.com/problems/thousand-separator)
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[中文文档](/solution/1500-1599/1556.Thousand Separator/README.md)
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## Description
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<p>Given an&nbsp;integer <code>n</code>, add a dot (&quot;.&quot;)&nbsp;as the thousands separator and return it in&nbsp;string format.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 987
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<strong>Output:</strong> &quot;987&quot;
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 1234
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<strong>Output:</strong> &quot;1.234&quot;
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> n = 123456789
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<strong>Output:</strong> &quot;123.456.789&quot;
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</pre>
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<p><strong>Example 4:</strong></p>
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<pre>
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<strong>Input:</strong> n = 0
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<strong>Output:</strong> &quot;0&quot;
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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;= n &lt; 2^31</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1500-1599/1557.Minimum Number of Vertices to Reach All Nodes/README.md

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# [1557. 可以到达所有点的最少点数目](https://leetcode-cn.com/problems/minimum-number-of-vertices-to-reach-all-nodes)
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[English Version](/solution/1500-1599/1557.Minimum Number of Vertices to Reach All Nodes/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个 <strong>有向无环图</strong>&nbsp;， <code>n</code>&nbsp;个节点编号为 <code>0</code>&nbsp;到 <code>n-1</code>&nbsp;，以及一个边数组 <code>edges</code>&nbsp;，其中 <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>]</code>&nbsp;表示一条从点&nbsp;&nbsp;<code>from<sub>i</sub></code>&nbsp;到点&nbsp;<code>to<sub>i</sub></code>&nbsp;的有向边。</p>
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<p>找到最小的点集使得从这些点出发能到达图中所有点。题目保证解存在且唯一。</p>
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<p>你可以以任意顺序返回这些节点编号。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/08/22/5480e1.png" style="height: 181px; width: 231px;"></p>
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<pre><strong>输入：</strong>n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]]
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<strong>输出：</strong>[0,3]
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<strong>解释：</strong>从单个节点出发无法到达所有节点。从 0 出发我们可以到达 [0,1,2,5] 。从 3 出发我们可以到达 [3,4,2,5] 。所以我们输出 [0,3] 。</pre>
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<p><strong>示例 2：</strong></p>
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<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/08/22/5480e2.png" style="height: 201px; width: 201px;"></p>
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<pre><strong>输入：</strong>n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]]
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<strong>输出：</strong>[0,2,3]
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<strong>解释：</strong>注意到节点 0，3 和 2 无法从其他节点到达，所以我们必须将它们包含在结果点集中，这些点都能到达节点 1 和 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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<ul>
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<li><code>2 &lt;= n &lt;= 10^5</code></li>
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<li><code>1 &lt;= edges.length &lt;= min(10^5, n * (n - 1) / 2)</code></li>
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<li><code>edges[i].length == 2</code></li>
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<li><code>0 &lt;= from<sub>i,</sub>&nbsp;to<sub>i</sub> &lt; n</code></li>
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<li>所有点对&nbsp;<code>(from<sub>i</sub>, to<sub>i</sub>)</code>&nbsp;互不相同。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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