feat: add solutions to lc problems: No.0767,2244

* No.0767.Reorganize String
* No.2244.Minimum Rounds to Complete All Tasks

Author: yanglbme

solution/0000-0099/0006.ZigZag Conversion/README.md solution/0000-0099/0006.Zigzag Conversion/README.md

@@ -1,6 +1,6 @@
 # [6. Z 字形变换](https://leetcode.cn/problems/zigzag-conversion)
 
-[English Version](/solution/0000-0099/0006.ZigZag%20Conversion/README_EN.md)
+[English Version](/solution/0000-0099/0006.Zigzag%20Conversion/README_EN.md)
 
 ## 题目描述
 
@@ -30,7 +30,6 @@ string convert(string s, int numRows);</pre>
 <strong>输入：</strong>s = "PAYPALISHIRING", numRows = 3
 <strong>输出：</strong>"PAHNAPLSIIGYIR"
 </pre>
-
 <strong>示例 2：</strong>
 
 <pre>
@@ -60,6 +59,7 @@ P     I
 	<li><code>1 <= numRows <= 1000</code></li>
 </ul>
 
+
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/0000-0099/0006.ZigZag Conversion/README_EN.md solution/0000-0099/0006.Zigzag Conversion/README_EN.md

@@ -1,6 +1,6 @@
-# [6. ZigZag Conversion](https://leetcode.com/problems/zigzag-conversion)
+# [6. Zigzag Conversion](https://leetcode.com/problems/zigzag-conversion)
 
-[中文文档](/solution/0000-0099/0006.ZigZag%20Conversion/README.md)
+[中文文档](/solution/0000-0099/0006.Zigzag%20Conversion/README.md)
 
 ## Description
 
@@ -56,6 +56,7 @@ P     I
 	<li><code>1 &lt;= numRows &lt;= 1000</code></li>
 </ul>
 
+
 ## Solutions
 
 <!-- tabs:start -->

solution/0000-0099/0006.ZigZag Conversion/Solution.cpp solution/0000-0099/0006.Zigzag Conversion/Solution.cpp

@@ -1,4 +1,4 @@
-# [146. LRU 缓存机制](https://leetcode.cn/problems/lru-cache)
+# [146. LRU 缓存](https://leetcode.cn/problems/lru-cache)
 
 [English Version](/solution/0100-0199/0146.LRU%20Cache/README_EN.md)
 
@@ -57,10 +57,13 @@ lRUCache.get(4);    // 返回 4
 	<li>最多调用 <code>2 * 10<sup>5</sup></code> 次 <code>get</code> 和 <code>put</code></li>
 </ul>
 
+
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：哈希表 + 双向链表**
+
 “哈希表 + 双向链表”实现。其中：
 
 -   双向链表按照被使用的顺序存储 kv 键值对，靠近头部的 kv 键值对是最近使用的，而靠近尾部的键值对是最久未使用的。

solution/0000-0099/0006.ZigZag Conversion/Solution.cs solution/0000-0099/0006.Zigzag Conversion/Solution.cs

@@ -49,6 +49,7 @@ lRUCache.get(4);    // return 4
 	<li>At most 2<code>&nbsp;* 10<sup>5</sup></code>&nbsp;calls will be made to <code>get</code> and <code>put</code>.</li>
 </ul>
 
+
 ## Solutions
 
 <!-- tabs:start -->

solution/0000-0099/0006.ZigZag Conversion/Solution.go solution/0000-0099/0006.Zigzag Conversion/Solution.go

@@ -1,6 +1,6 @@
 # [158. 用 Read4 读取 N 个字符 II](https://leetcode.cn/problems/read-n-characters-given-read4-ii-call-multiple-times)
 
-[English Version](/solution/0100-0199/0158.Read%20N%20Characters%20Given%20Read4%20II%20-%20Call%20multiple%20times/README_EN.md)
+[English Version](/solution/0100-0199/0158.Read%20N%20Characters%20Given%20read4%20II%20-%20Call%20Multiple%20Times/README_EN.md)
 
 ## 题目描述
 
@@ -56,7 +56,7 @@ read4(buf4); // read4 返回 0。现在 buf4 = ""，fp 指向文件末尾</code>
 <ul>
 	<li>你 <strong>不能</strong> 直接操作该文件，文件只能通过 <code>read4</code> 获取而 <strong>不能</strong> 通过 <code>read</code>。</li>
 	<li><code>read</code>&nbsp; 函数可以被调用&nbsp;<strong>多次</strong>。</li>
-	<li>请记得&nbsp;<strong>重置&nbsp;</strong>在 Solution 中声明的类变量（静态变量），因为类变量会&nbsp;<strong>在多个测试用例中保持不变</strong>，影响判题准确。请 <a href="https://support.leetcode.cn/hc/kb/section/1071534/" target="_blank">查阅</a> 这里。</li>
+	<li>请记得&nbsp;<strong>重置&nbsp;</strong>在 Solution 中声明的类变量（静态变量），因为类变量会&nbsp;<strong>在多个测试用例中保持不变</strong>，影响判题准确。请 <a href="https://support.leetcode-cn.com/hc/kb/section/1071534/" target="_blank">查阅</a> 这里。</li>
 	<li>你可以假定目标缓存数组&nbsp;<code>buf</code> 保证有足够的空间存下 n 个字符。&nbsp;</li>
 	<li>保证在一个给定测试用例中，<code>read</code> 函数使用的是同一个 <code>buf</code>。</li>
 </ul>
@@ -102,6 +102,7 @@ sol.read (buf, 1); // 我们已经到达文件的末尾，不能读取更多的
 	<li><code>1 &lt;= queries[i] &lt;= 500</code></li>
 </ul>
 
+
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
@@ -124,6 +125,12 @@ sol.read (buf, 1); // 我们已经到达文件的末尾，不能读取更多的
 
 ```
 
+### **TypeScript**
+
+```ts
+
+```
+
 ### **...**
 
 ```

solution/0000-0099/0006.ZigZag Conversion/Solution.java solution/0000-0099/0006.Zigzag Conversion/Solution.java

@@ -1,6 +1,6 @@
-# [158. Read N Characters Given Read4 II - Call multiple times](https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times)
+# [158. Read N Characters Given read4 II - Call Multiple Times](https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times)
 
-[中文文档](/solution/0100-0199/0158.Read%20N%20Characters%20Given%20Read4%20II%20-%20Call%20multiple%20times/README.md)
+[中文文档](/solution/0100-0199/0158.Read%20N%20Characters%20Given%20read4%20II%20-%20Call%20Multiple%20Times/README.md)
 
 ## Description
 
@@ -97,6 +97,7 @@ sol.read(buf, 1); // We have reached the end of file, no more characters can be
 	<li><code>1 &lt;= queries[i] &lt;= 500</code></li>
 </ul>
 
+
 ## Solutions
 
 <!-- tabs:start -->
@@ -113,6 +114,12 @@ sol.read(buf, 1); // We have reached the end of file, no more characters can be
 
 ```
 
+### **TypeScript**
+
+```ts
+
+```
+
 ### **...**
 
 ```

solution/0000-0099/0006.ZigZag Conversion/Solution.js solution/0000-0099/0006.Zigzag Conversion/Solution.js

@@ -39,22 +39,170 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：哈希表**
+
+利用哈希表 cnt 统计字符串 s 中每个字符出现的次数。
+
+若最大的出现次数 mx 大于 `(n + 1) / 2`，说明一定会存在两个相同字符相邻，直接返回 ''。
+
+否则，按字符出现频率从大到小遍历，依次间隔 1 个位置填充字符。若位置大于等于 n，则重置为 1 继续填充。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def reorganizeString(self, s: str) -> str:
+        n = len(s)
+        cnt = Counter(s)
+        mx = max(cnt.values())
+        if mx > (n + 1) // 2:
+            return ''
+        i = 0
+        ans = [None] * n
+        for k, v in cnt.most_common():
+            while v:
+                ans[i] = k
+                v -= 1
+                i += 2
+                if i >= n:
+                    i = 1
+        return ''.join(ans)
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public String reorganizeString(String s) {
+        int[] cnt = new int[26];
+        int mx = 0;
+        for (char c : s.toCharArray()) {
+            int t = c - 'a';
+            ++cnt[t];
+            mx = Math.max(mx, cnt[t]);
+        }
+        int n = s.length();
+        if (mx > (n + 1) / 2) {
+            return "";
+        }
+        int k = 0;
+        for (int v : cnt) {
+            if (v > 0) {
+                ++k;
+            }
+        }
+        int[][] m = new int[k][2];
+        k = 0;
+        for (int i = 0; i < 26; ++i) {
+            if (cnt[i] > 0) {
+                m[k++] = new int[]{cnt[i], i};
+            }
+        }
+        Arrays.sort(m, (a, b) -> b[0] - a[0]);
+        k = 0;
+        StringBuilder ans = new StringBuilder(s);
+        for (int[] e : m) {
+            int v = e[0], i = e[1];
+            while (v-- > 0) {
+                ans.setCharAt(k, (char) ('a' + i));
+                k += 2;
+                if (k >= n) {
+                    k = 1;
+                }
+            }
+        }
+        return ans.toString();
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    string reorganizeString(string s) {
+        vector<int> cnt(26);
+        for (char& c : s) ++cnt[c - 'a'];
+        int mx = *max_element(cnt.begin(), cnt.end());
+        int n = s.size();
+        if (mx > (n + 1) / 2) return "";
+        vector<vector<int>> m;
+        for (int i = 0; i < 26; ++i)
+        {
+            if (cnt[i]) m.push_back({cnt[i], i});
+        }
+        sort(m.begin(), m.end());
+        reverse(m.begin(), m.end());
+        string ans = s;
+        int k = 0;
+        for (auto& e : m)
+        {
+            int v = e[0], i = e[1];
+            while (v--)
+            {
+                ans[k] = 'a' + i;
+                k += 2;
+                if (k >= n) k = 1;
+            }
+        }
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+func reorganizeString(s string) string {
+	cnt := make([]int, 26)
+	mx := 0
+	for _, c := range s {
+		t := c - 'a'
+		cnt[t]++
+		mx = max(mx, cnt[t])
+	}
+	n := len(s)
+	if mx > (n+1)/2 {
+		return ""
+	}
+	m := [][]int{}
+	for i, v := range cnt {
+		if v > 0 {
+			m = append(m, []int{v, i})
+		}
+	}
+	sort.Slice(m, func(i, j int) bool {
+		return m[i][0] > m[j][0]
+	})
+	ans := make([]byte, n)
+	k := 0
+	for _, e := range m {
+		v, i := e[0], e[1]
+		for v > 0 {
+			ans[k] = byte('a' + i)
+			k += 2
+			if k >= n {
+				k = 1
+			}
+			v--
+		}
+	}
+	return string(ans)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
 ```
 
 ### **...**