feat: add new lc problems

Author: yanglbme

solution/1100-1199/1129.Shortest Path with Alternating Colors/README.md

@@ -6,9 +6,14 @@
 
 <!-- 这里写题目描述 -->
 
-<p>在一个有向图中，节点分别标记为&nbsp;<code>0, 1, ..., n-1</code>。图中每条边为红色或者蓝色，且存在自环或平行边。</p>
+<p>给定一个整数 <code>n</code>，即有向图中的节点数，其中节点标记为 <code>0</code> 到 <code>n - 1</code>。图中的每条边为红色或者蓝色，并且可能存在自环或平行边。</p>
 
-<p><code>red_edges</code>&nbsp;中的每一个&nbsp;<code>[i, j]</code>&nbsp;对表示从节点 <code>i</code> 到节点 <code>j</code> 的红色有向边。类似地，<code>blue_edges</code>&nbsp;中的每一个&nbsp;<code>[i, j]</code>&nbsp;对表示从节点 <code>i</code> 到节点 <code>j</code> 的蓝色有向边。</p>
+<p>给定两个数组&nbsp;<code>redEdges</code>&nbsp;和&nbsp;<code>blueEdges</code>，其中：</p>
+
+<ul>
+	<li><code>redEdges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code>&nbsp;表示图中存在一条从节点&nbsp;<code>a<sub>i</sub></code>&nbsp;到节点&nbsp;<code>b<sub>i</sub></code>&nbsp;的红色有向边，</li>
+	<li><code>blueEdges[j] = [u<sub>j</sub>, v<sub>j</sub>]</code>&nbsp;表示图中存在一条从节点&nbsp;<code>u<sub>j</sub></code>&nbsp;到节点&nbsp;<code>v<sub>j</sub></code>&nbsp;的蓝色有向边。</li>
+</ul>
 
 <p>返回长度为 <code>n</code> 的数组&nbsp;<code>answer</code>，其中&nbsp;<code>answer[X]</code>&nbsp;是从节点&nbsp;<code>0</code>&nbsp;到节点&nbsp;<code>X</code>&nbsp;的红色边和蓝色边交替出现的最短路径的长度。如果不存在这样的路径，那么 <code>answer[x] = -1</code>。</p>
 
@@ -28,37 +33,15 @@
 <strong>输出：</strong>[0,1,-1]
 </pre>
 
-<p><strong>示例 3：</strong></p>
-
-<pre>
-<strong>输入：</strong>n = 3, red_edges = [[1,0]], blue_edges = [[2,1]]
-<strong>输出：</strong>[0,-1,-1]
-</pre>
-
-<p><strong>示例 4：</strong></p>
-
-<pre>
-<strong>输入：</strong>n = 3, red_edges = [[0,1]], blue_edges = [[1,2]]
-<strong>输出：</strong>[0,1,2]
-</pre>
-
-<p><strong>示例 5：</strong></p>
-
-<pre>
-<strong>输入：</strong>n = 3, red_edges = [[0,1],[0,2]], blue_edges = [[1,0]]
-<strong>输出：</strong>[0,1,1]
-</pre>
-
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
 	<li><code>1 &lt;= n &lt;= 100</code></li>
-	<li><code>red_edges.length &lt;= 400</code></li>
-	<li><code>blue_edges.length &lt;= 400</code></li>
-	<li><code>red_edges[i].length == blue_edges[i].length == 2</code></li>
-	<li><code>0 &lt;= red_edges[i][j], blue_edges[i][j] &lt; n</code></li>
+	<li><code>0 &lt;= redEdges.length,&nbsp;blueEdges.length &lt;= 400</code></li>
+	<li><code>redEdges[i].length == blueEdges[j].length == 2</code></li>
+	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub>, u<sub>j</sub>, v<sub>j</sub>&nbsp;&lt; n</code></li>
 </ul>
 
 ## 解法

solution/2500-2599/2566.Maximum Difference by Remapping a Digit/README.md

@@ -0,0 +1,159 @@
+# [2566. 替换一个数字后的最大差值](https://leetcode.cn/problems/maximum-difference-by-remapping-a-digit)
+
+[English Version](/solution/2500-2599/2566.Maximum%20Difference%20by%20Remapping%20a%20Digit/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个整数&nbsp;<code>num</code>&nbsp;。你知道 Danny Mittal 会偷偷将 <code>0</code>&nbsp;到 <code>9</code>&nbsp;中的一个数字 <strong>替换</strong> 成另一个数字。</p>
+
+<p>请你返回将 <code>num</code>&nbsp;中&nbsp;<strong>恰好一个</strong>&nbsp;数字进行替换后，得到的最大值和最小值的差位多少。</p>
+
+<p><strong>注意：</strong></p>
+
+<ul>
+	<li>当 Danny 将一个数字 <code>d1</code> 替换成另一个数字 <code>d2</code> 时，Danny 需要将&nbsp;<code>nums</code>&nbsp;中所有 <code>d1</code>&nbsp;都替换成&nbsp;<code>d2</code>&nbsp;。</li>
+	<li>Danny 可以将一个数字替换成它自己，也就是说&nbsp;<code>num</code>&nbsp;可以不变。</li>
+	<li>Danny 可以将数字分别替换成两个不同的数字分别得到最大值和最小值。</li>
+	<li>替换后得到的数字可以包含前导 0 。</li>
+	<li>Danny Mittal 获得周赛 326 前 10 名，让我们恭喜他。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>num = 11891
+<b>输出：</b>99009
+<b>解释：</b>
+为了得到最大值，我们将数字 1 替换成数字 9 ，得到 99899 。
+为了得到最小值，我们将数字 1 替换成数字 0 ，得到 890 。
+两个数字的差值为 99009 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>num = 90
+<b>输出：</b>99
+<strong>解释：</strong>
+可以得到的最大值是 99（将 0 替换成 9），最小值是 0（将 9 替换成 0）。
+所以我们得到 99 。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= num &lt;= 10<sup>8</sup></code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def minMaxDifference(self, num: int) -> int:
+        s = str(num)
+        mi = int(s.replace(s[0], '0'))
+        for c in s:
+            if c != '9':
+                return int(s.replace(c, '9')) - mi
+        return num - mi
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int minMaxDifference(int num) {
+        String s = String.valueOf(num);
+        int mi = Integer.parseInt(s.replace(s.charAt(0), '0'));
+        for (char c : s.toCharArray()) {
+            if (c != '9') {
+                return Integer.parseInt(s.replace(c, '9')) - mi;
+            }
+        }
+        return num - mi;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minMaxDifference(int num) {
+        string s = to_string(num);
+        string t = s;
+        char first = s[0];
+        for (char& c : s) {
+            if (c == first) {
+                c = '0';
+            }
+        }
+        int mi = stoi(s);
+        for (int i = 0; i < t.size(); ++i) {
+            if (t[i] != '9') {
+                char second = t[i];
+                for (int j = i; j < t.size(); ++j) {
+                    if (t[j] == second) {
+                        t[j] = '9';
+                    }
+                }
+                return stoi(t) - mi;
+            }
+        }
+        return num - mi;
+    }
+};
+```
+
+### **Go**
+
+```go
+func minMaxDifference(num int) int {
+	s := []byte(strconv.Itoa(num))
+	first := s[0]
+	for i := range s {
+		if s[i] == first {
+			s[i] = '0'
+		}
+	}
+	mi, _ := strconv.Atoi(string(s))
+	t := []byte(strconv.Itoa(num))
+	for i := range t {
+		if t[i] != '9' {
+			second := t[i]
+			for j := i; j < len(t); j++ {
+				if t[j] == second {
+					t[j] = '9'
+				}
+			}
+			mx, _ := strconv.Atoi(string(t))
+			return mx - mi
+		}
+	}
+	return num - mi
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2500-2599/2566.Maximum Difference by Remapping a Digit/README_EN.md

@@ -0,0 +1,149 @@
+# [2566. Maximum Difference by Remapping a Digit](https://leetcode.com/problems/maximum-difference-by-remapping-a-digit)
+
+[中文文档](/solution/2500-2599/2566.Maximum%20Difference%20by%20Remapping%20a%20Digit/README.md)
+
+## Description
+
+<p>You are given an integer <code>num</code>. You know that Danny Mittal will sneakily <strong>remap</strong> one of the <code>10</code> possible digits (<code>0</code> to <code>9</code>) to another digit.</p>
+
+<p>Return <em>the difference between the maximum and minimum</em><em>&nbsp;values Danny can make by remapping&nbsp;<strong>exactly</strong> <strong>one</strong> digit</em><em> in </em><code>num</code>.</p>
+
+<p><strong>Notes:</strong></p>
+
+<ul>
+	<li>When Danny remaps a digit <font face="monospace">d1</font>&nbsp;to another digit <font face="monospace">d2</font>, Danny replaces all occurrences of <code>d1</code>&nbsp;in <code>num</code>&nbsp;with <code>d2</code>.</li>
+	<li>Danny can remap a digit to itself, in which case <code>num</code>&nbsp;does not change.</li>
+	<li>Danny can remap different digits for obtaining minimum and maximum values respectively.</li>
+	<li>The resulting number after remapping can contain leading zeroes.</li>
+	<li>We mentioned &quot;Danny Mittal&quot; to congratulate him on being in the top 10 in Weekly Contest 326.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> num = 11891
+<strong>Output:</strong> 99009
+<strong>Explanation:</strong> 
+To achieve the maximum value, Danny can remap the digit 1 to the digit 9 to yield 99899.
+To achieve the minimum value, Danny can remap the digit 1 to the digit 0, yielding 890.
+The difference between these two numbers is 99009.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> num = 90
+<strong>Output:</strong> 99
+<strong>Explanation:</strong>
+The maximum value that can be returned by the function is 99 (if 0 is replaced by 9) and the minimum value that can be returned by the function is 0 (if 9 is replaced by 0).
+Thus, we return 99.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= num &lt;= 10<sup>8</sup></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def minMaxDifference(self, num: int) -> int:
+        s = str(num)
+        mi = int(s.replace(s[0], '0'))
+        for c in s:
+            if c != '9':
+                return int(s.replace(c, '9')) - mi
+        return num - mi
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int minMaxDifference(int num) {
+        String s = String.valueOf(num);
+        int mi = Integer.parseInt(s.replace(s.charAt(0), '0'));
+        for (char c : s.toCharArray()) {
+            if (c != '9') {
+                return Integer.parseInt(s.replace(c, '9')) - mi;
+            }
+        }
+        return num - mi;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minMaxDifference(int num) {
+        string s = to_string(num);
+        string t = s;
+        char first = s[0];
+        for (char& c : s) {
+            if (c == first) {
+                c = '0';
+            }
+        }
+        int mi = stoi(s);
+        for (int i = 0; i < t.size(); ++i) {
+            if (t[i] != '9') {
+                char second = t[i];
+                for (int j = i; j < t.size(); ++j) {
+                    if (t[j] == second) {
+                        t[j] = '9';
+                    }
+                }
+                return stoi(t) - mi;
+            }
+        }
+        return num - mi;
+    }
+};
+```
+
+### **Go**
+
+```go
+func minMaxDifference(num int) int {
+	s := []byte(strconv.Itoa(num))
+	first := s[0]
+	for i := range s {
+		if s[i] == first {
+			s[i] = '0'
+		}
+	}
+	mi, _ := strconv.Atoi(string(s))
+	t := []byte(strconv.Itoa(num))
+	for i := range t {
+		if t[i] != '9' {
+			second := t[i]
+			for j := i; j < len(t); j++ {
+				if t[j] == second {
+					t[j] = '9'
+				}
+			}
+			mx, _ := strconv.Atoi(string(t))
+			return mx - mi
+		}
+	}
+	return num - mi
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2500-2599/2566.Maximum Difference by Remapping a Digit/Solution.cpp

@@ -0,0 +1,26 @@
+class Solution {
+public:
+    int minMaxDifference(int num) {
+        string s = to_string(num);
+        string t = s;
+        char first = s[0];
+        for (char& c : s) {
+            if (c == first) {
+                c = '0';
+            }
+        }
+        int mi = stoi(s);
+        for (int i = 0; i < t.size(); ++i) {
+            if (t[i] != '9') {
+                char second = t[i];
+                for (int j = i; j < t.size(); ++j) {
+                    if (t[j] == second) {
+                        t[j] = '9';
+                    }
+                }
+                return stoi(t) - mi;
+            }
+        }
+        return num - mi;
+    }
+};