feat: add solutions to lc problems: No.216,247,249 (doocs#2634)

Author: yanglbme

solution/0200-0299/0216.Combination Sum III/README.md

@@ -71,16 +71,6 @@
 -   如果 $i \gt 9$ 或者 $i \gt s$ 或者当前搜索路径 $t$ 的长度大于 $k$，说明当前搜索路径不可能是答案，直接返回。
 -   否则，我们可以选择将数字 $i$ 加入搜索路径 $t$ 中，然后继续搜索，即执行 $dfs(i + 1, s - i)$，搜索完成后，将 $i$ 从搜索路径 $t$ 中移除；我们也可以选择不将数字 $i$ 加入搜索路径 $t$ 中，直接执行 $dfs(i + 1, s)$。
 
-方式二：
-
--   如果 $s = 0$，且当前搜索路径 $t$ 的长度为 $k$，说明找到了一组答案，将 $t$ 加入 $ans$ 中，然后返回。
--   如果 $i \gt 9$ 或者 $i \gt s$ 或者当前搜索路径 $t$ 的长度大于 $k$，说明当前搜索路径不可能是答案，直接返回。
--   否则，我们枚举下一个数字 $j$，即 $j \in [i, 9]$，将数字 $j$ 加入搜索路径 $t$ 中，然后继续搜索，即执行 $dfs(j + 1, s - j)$，搜索完成后，将 $j$ 从搜索路径 $t$ 中移除。
-
-在主函数中，我们调用 $dfs(1, n)$，即从数字 $1$ 开始枚举，剩下和为 $n$ 的数字需要枚举。搜索完成后，即可得到所有的答案。
-
-时间复杂度 $(C_{9}^k \times k)$，空间复杂度 $O(k)$。
-
 <!-- tabs:start -->
 
 ```python
@@ -280,19 +270,15 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### 方法二：二进制枚举
-
-我们可以用一个长度为 $9$ 的二进制整数表示数字 $1$ 到 $9$ 的选取情况，其中二进制整数的第 $i$ 位表示数字 $i + 1$ 是否被选取，如果第 $i$ 位为 $1$，则表示数字 $i + 1$ 被选取，否则表示数字 $i + 1$ 没有被选取。
-
-我们在 $[0, 2^9)$ 范围内枚举二进制整数，对于当前枚举到的二进制整数 $mask$，如果 $mask$ 的二进制表示中 $1$ 的个数为 $k$，且 $mask$ 的二进制表示中 $1$ 所对应的数字之和为 $n$，则说明 $mask$ 对应的数字选取方案是一组答案。我们将 $mask$ 对应的数字选取方案加入答案即可。
+方式二：
 
-时间复杂度 $O(2^9 \times 9)$，空间复杂度 $O(k)$。
+-   如果 $s = 0$，且当前搜索路径 $t$ 的长度为 $k$，说明找到了一组答案，将 $t$ 加入 $ans$ 中，然后返回。
+-   如果 $i \gt 9$ 或者 $i \gt s$ 或者当前搜索路径 $t$ 的长度大于 $k$，说明当前搜索路径不可能是答案，直接返回。
+-   否则，我们枚举下一个数字 $j$，即 $j \in [i, 9]$，将数字 $j$ 加入搜索路径 $t$ 中，然后继续搜索，即执行 $dfs(j + 1, s - j)$，搜索完成后，将 $j$ 从搜索路径 $t$ 中移除。
 
-相似题目：
+在主函数中，我们调用 $dfs(1, n)$，即从数字 $1$ 开始枚举，剩下和为 $n$ 的数字需要枚举。搜索完成后，即可得到所有的答案。
 
--   [39. 组合总和](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0039.Combination%20Sum/README.md)
--   [40. 组合总和 II](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0040.Combination%20Sum%20II/README.md)
--   [77. 组合](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0077.Combinations/README.md)
+时间复杂度 $(C_{9}^k \times k)$，空间复杂度 $O(k)$。
 
 <!-- tabs:start -->
 
@@ -459,7 +445,19 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### 方法三
+### 方法二：二进制枚举
+
+我们可以用一个长度为 $9$ 的二进制整数表示数字 $1$ 到 $9$ 的选取情况，其中二进制整数的第 $i$ 位表示数字 $i + 1$ 是否被选取，如果第 $i$ 位为 $1$，则表示数字 $i + 1$ 被选取，否则表示数字 $i + 1$ 没有被选取。
+
+我们在 $[0, 2^9)$ 范围内枚举二进制整数，对于当前枚举到的二进制整数 $mask$，如果 $mask$ 的二进制表示中 $1$ 的个数为 $k$，且 $mask$ 的二进制表示中 $1$ 所对应的数字之和为 $n$，则说明 $mask$ 对应的数字选取方案是一组答案。我们将 $mask$ 对应的数字选取方案加入答案即可。
+
+时间复杂度 $O(2^9 \times 9)$，空间复杂度 $O(k)$。
+
+相似题目：
+
+-   [39. 组合总和](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0039.Combination%20Sum/README.md)
+-   [40. 组合总和 II](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0040.Combination%20Sum%20II/README.md)
+-   [77. 组合](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0077.Combinations/README.md)
 
 <!-- tabs:start -->
 

solution/0200-0299/0216.Combination Sum III/README_EN.md

@@ -56,7 +56,17 @@ Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Pruning + Backtracking (Two Approaches)
+
+We design a function $dfs(i, s)$, which represents that we are currently enumerating the number $i$, and there are still numbers with a sum of $s$ to be enumerated. The current search path is $t$, and the answer is $ans$.
+
+The execution logic of the function $dfs(i, s)$ is as follows:
+
+Approach One:
+
+-   If $s = 0$ and the length of the current search path $t$ is $k$, it means that a group of answers has been found. Add $t$ to $ans$ and then return.
+-   If $i \gt 9$ or $i \gt s$ or the length of the current search path $t$ is greater than $k$, it means that the current search path cannot be the answer, so return directly.
+-   Otherwise, we can choose to add the number $i$ to the search path $t$, and then continue to search, i.e., execute $dfs(i + 1, s - i)$. After the search is completed, remove $i$ from the search path $t$; we can also choose not to add the number $i$ to the search path $t$, and directly execute $dfs(i + 1, s)$.
 
 <!-- tabs:start -->
 
@@ -257,7 +267,15 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### Solution 2
+Another approach:
+
+-   If $s = 0$ and the length of the current search path $t$ is $k$, it means that a group of answers has been found. Add $t$ to $ans$ and then return.
+-   If $i \gt 9$ or $i \gt s$ or the length of the current search path $t$ is greater than $k$, it means that the current search path cannot be the answer, so return directly.
+-   Otherwise, we enumerate the next number $j$, i.e., $j \in [i, 9]$, add the number $j$ to the search path $t$, and then continue to search, i.e., execute $dfs(j + 1, s - j)$. After the search is completed, remove $j$ from the search path $t$.
+
+In the main function, we call $dfs(1, n)$, i.e., start enumerating from the number $1$, and the remaining numbers with a sum of $n$ need to be enumerated. After the search is completed, we can get all the answers.
+
+The time complexity is $(C_{9}^k \times k)$, and the space complexity is $O(k)$.
 
 <!-- tabs:start -->
 
@@ -424,7 +442,19 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### Solution 3
+### Solution 2: Binary Enumeration
+
+We can use a binary integer of length $9$ to represent the selection of numbers $1$ to $9$, where the $i$-th bit of the binary integer represents whether the number $i + 1$ is selected. If the $i$-th bit is $1$, it means that the number $i + 1$ is selected, otherwise, it means that the number $i + 1$ is not selected.
+
+We enumerate binary integers in the range of $[0, 2^9)$. For the currently enumerated binary integer $mask$, if the number of $1$s in the binary representation of $mask$ is $k$, and the sum of the numbers corresponding to $1$ in the binary representation of $mask$ is $n$, it means that the number selection scheme corresponding to $mask$ is a group of answers. We can add the number selection scheme corresponding to $mask$ to the answer.
+
+The time complexity is $O(2^9 \times 9)$, and the space complexity is $O(k)$.
+
+Similar problems:
+
+-   [39. Combination Sum](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0039.Combination%20Sum/README_EN.md)
+-   [40. Combination Sum II](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0040.Combination%20Sum%20II/README_EN.md)
+-   [77. Combinations](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0077.Combinations/README_EN.md)
 
 <!-- tabs:start -->
 

solution/0200-0299/0246.Strobogrammatic Number/README.md

@@ -42,9 +42,9 @@
 
 我们定义一个数组 $d$，其中 $d[i]$ 表示数字 $i$ 旋转 180° 之后的数字。如果 $d[i]$ 为 $-1$，表示数字 $i$ 不能旋转 180° 得到一个数字。
 
-定义两个指针 $i$ 和 $j$，分别指向字符串的左右两端，然后不断向中间移动指针，判断 $d[num[i]]$ 和 $num[j]$ 是否相等，如果不相等，说明该字符串不是中心对称数 🔒，直接返回 $false$ 即可。如果 $i \gt j$，说明遍历完了字符串，返回 $true$。
+定义两个指针 $i$ 和 $j$，分别指向字符串的左右两端，然后不断向中间移动指针，判断 $d[num[i]]$ 和 $num[j]$ 是否相等，如果不相等，说明该字符串不是中心对称数，直接返回 $false$ 即可。如果 $i \gt j$，说明遍历完了字符串，返回 $true$。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串的长度。
+时间复杂度 $O(n)$，其中 $n$ 为字符串的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 

solution/0200-0299/0246.Strobogrammatic Number/README_EN.md

@@ -43,7 +43,13 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Two Pointers Simulation
+
+We define an array $d$, where $d[i]$ represents the number after rotating the digit $i$ by 180°. If $d[i]$ is $-1$, it means that the digit $i$ cannot be rotated 180° to get a valid digit.
+
+We define two pointers $i$ and $j$, pointing to the left and right ends of the string, respectively. Then we continuously move the pointers towards the center, checking whether $d[num[i]]$ and $num[j]$ are equal. If they are not equal, it means that the string is not a strobogrammatic number, and we can directly return $false$. If $i > j$, it means that we have traversed the entire string, and we return $true$.
+
+The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 

solution/0200-0299/0247.Strobogrammatic Number II/README.md

@@ -55,7 +55,7 @@
 
 相似题目：
 
--   [248. 中心对称数 II 🔒I](https://github.com/doocs/leetcode/blob/main/solution/0200-0299/0248.Strobogrammatic%20Number%20III/README.md)
+-   [248. 中心对称数 III 🔒](https://github.com/doocs/leetcode/blob/main/solution/0200-0299/0248.Strobogrammatic%20Number%20III/README.md)
 
 <!-- tabs:start -->
 

solution/0200-0299/0247.Strobogrammatic Number II/README_EN.md

@@ -27,7 +27,25 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Recursion
+
+If the length is $1$, then the strobogrammatic numbers are only $0, 1, 8$; if the length is $2$, then the strobogrammatic numbers are only $11, 69, 88, 96$.
+
+We design a recursive function $dfs(u)$, which returns the strobogrammatic numbers of length $u$. The answer is $dfs(n)$.
+
+If $u$ is $0$, return a list containing an empty string, i.e., `[""]`; if $u$ is $1$, return the list `["0", "1", "8"]`.
+
+If $u$ is greater than $1$, we traverse all the strobogrammatic numbers of length $u - 2$. For each strobogrammatic number $v$, we add $1, 8, 6, 9$ to both sides of it, and we can get the strobogrammatic numbers of length `u`.
+
+Note that if $u \neq n$, we can also add $0$ to both sides of the strobogrammatic number.
+
+Finally, we return all the strobogrammatic numbers of length $n$.
+
+The time complexity is $O(2^{n+2})$.
+
+Similar problems:
+
+-   [248. Strobogrammatic Number III 🔒](https://github.com/doocs/leetcode/blob/main/solution/0200-0299/0248.Strobogrammatic%20Number%20III/README_EN.md)
 
 <!-- tabs:start -->
 

solution/0200-0299/0248.Strobogrammatic Number III/README_EN.md

@@ -30,7 +30,27 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Recursion
+
+If the length is $1$, then the strobogrammatic numbers are only $0, 1, 8$; if the length is $2$, then the strobogrammatic numbers are only $11, 69, 88, 96$.
+
+We design a recursive function $dfs(u)$, which returns the strobogrammatic numbers of length $u$.
+
+If $u$ is $0$, return a list containing an empty string, i.e., `[""]`; if $u$ is $1$, return the list `["0", "1", "8"]`.
+
+If $u$ is greater than $1$, we traverse all the strobogrammatic numbers of length $u - 2$. For each strobogrammatic number $v$, we add $1, 8, 6, 9$ to both sides of it, and we can get the strobogrammatic numbers of length $u$.
+
+Note that if $u \neq n$, we can also add $0$ to both sides of the strobogrammatic number.
+
+Let the lengths of $low$ and $high$ be $a$ and $b$ respectively.
+
+Next, we traverse all lengths in the range $[a,..b]$. For each length $n$, we get all strobogrammatic numbers $dfs(n)$, and then check whether they are in the range $[low, high]$. If they are, we increment the answer.
+
+The time complexity is $O(2^{n+2} \times \log n)$.
+
+Similar problems:
+
+-   [247. Strobogrammatic Number II](https://github.com/doocs/leetcode/blob/main/solution/0200-0299/0247.Strobogrammatic%20Number%20II/README_EN.md)
 
 <!-- tabs:start -->
 

solution/0200-0299/0249.Group Shifted Strings/README.md

@@ -30,45 +30,50 @@
 
 ## 解法
 
-### 方法一
+### 方法一：哈希表
+
+我们用一个哈希表 $g$ 来存储每个字符串移位后且首位为 '`a`' 的字符串。即 $g[t]$ 表示所有字符串移位后字符串为 $t$ 的字符串集合。
+
+我们遍历每个字符串，对于每个字符串，我们计算其移位后的字符串 $t$，然后将其加入到 $g[t]$ 中。
+
+最后，我们将 $g$ 中的所有值取出来，即为答案。
+
+时间复杂度 $O(L)$，空间复杂度 $O(L)$，其中 $L$ 为所有字符串的长度之和。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def groupStrings(self, strings: List[str]) -> List[List[str]]:
-        mp = defaultdict(list)
+        g = defaultdict(list)
         for s in strings:
+            diff = ord(s[0]) - ord("a")
             t = []
-            diff = ord(s[0]) - ord('a')
             for c in s:
-                d = ord(c) - diff
-                if d < ord('a'):
-                    d += 26
-                t.append(chr(d))
-            k = ''.join(t)
-            mp[k].append(s)
-        return list(mp.values())
+                c = ord(c) - diff
+                if c < ord("a"):
+                    c += 26
+                t.append(chr(c))
+            g["".join(t)].append(s)
+        return list(g.values())
 ```
 
 ```java
 class Solution {
     public List<List<String>> groupStrings(String[] strings) {
-        Map<String, List<String>> mp = new HashMap<>();
-        for (String s : strings) {
-            int diff = s.charAt(0) - 'a';
+        Map<String, List<String>> g = new HashMap<>();
+        for (var s : strings) {
             char[] t = s.toCharArray();
+            int diff = t[0] - 'a';
             for (int i = 0; i < t.length; ++i) {
-                char d = (char) (t[i] - diff);
-                if (d < 'a') {
-                    d += 26;
+                t[i] = (char) (t[i] - diff);
+                if (t[i] < 'a') {
+                    t[i] += 26;
                 }
-                t[i] = d;
             }
-            String key = new String(t);
-            mp.computeIfAbsent(key, k -> new ArrayList<>()).add(s);
+            g.computeIfAbsent(new String(t), k -> new ArrayList<>()).add(s);
         }
-        return new ArrayList<>(mp.values());
+        return new ArrayList<>(g.values());
     }
 }
 ```
@@ -77,38 +82,44 @@ class Solution {
 class Solution {
 public:
     vector<vector<string>> groupStrings(vector<string>& strings) {
-        unordered_map<string, vector<string>> mp;
+        unordered_map<string, vector<string>> g;
         for (auto& s : strings) {
+            string t;
             int diff = s[0] - 'a';
-            string t = s;
-            for (int i = 0; i < t.size(); ++i) {
-                char d = t[i] - diff;
-                if (d < 'a') d += 26;
-                t[i] = d;
+            for (int i = 0; i < s.size(); ++i) {
+                char c = s[i] - diff;
+                if (c < 'a') {
+                    c += 26;
+                }
+                t.push_back(c);
             }
-            cout << t << endl;
-            mp[t].push_back(s);
+            g[t].emplace_back(s);
         }
         vector<vector<string>> ans;
-        for (auto& e : mp)
-            ans.push_back(e.second);
+        for (auto& p : g) {
+            ans.emplace_back(move(p.second));
+        }
         return ans;
     }
 };
 ```
 
 ```go
 func groupStrings(strings []string) [][]string {
-	mp := make(map[string][]string)
+	g := make(map[string][]string)
 	for _, s := range strings {
-		k := ""
-		for i := range s {
-			k += string((s[i]-s[0]+26)%26 + 'a')
+		t := []byte(s)
+		diff := t[0] - 'a'
+		for i := range t {
+			t[i] -= diff
+			if t[i] < 'a' {
+				t[i] += 26
+			}
 		}
-		mp[k] = append(mp[k], s)
+		g[string(t)] = append(g[string(t)], s)
 	}
-	var ans [][]string
-	for _, v := range mp {
+	ans := make([][]string, 0, len(g))
+	for _, v := range g {
 		ans = append(ans, v)
 	}
 	return ans