feat: add solutions to lc problem: No.3167

lcof/面试题56 - II. 数组中数字出现的次数 II/README.md

@@ -208,22 +208,22 @@ public class Solution {
 class Solution {
     func singleNumber(_ nums: [Int]) -> Int {
         var bitCounts = [Int](repeating: 0, count: 32)
-        
+
         for num in nums {
             var x = num
             for i in 0..<32 {
                 bitCounts[i] += x & 1
                 x >>= 1
             }
         }
-        
+
         var result = 0
         for i in 0..<32 {
             if bitCounts[i] % 3 == 1 {
                 result |= 1 << i
             }
         }
-        
+
         return result
     }
 }

lcof/面试题57 - II. 和为s的连续正数序列/README.md

@@ -218,7 +218,7 @@ class Solution {
     func findContinuousSequence(_ target: Int) -> [[Int]] {
         var l = 1, r = 2
         var result = [[Int]]()
-        
+
         while l < r {
             let sum = (l + r) * (r - l + 1) / 2
             if sum == target {
@@ -234,7 +234,7 @@ class Solution {
                 l += 1
             }
         }
-        
+
         return result
     }
 }

lcof/面试题57. 和为s的两个数字/README.md

@@ -219,7 +219,7 @@ class Solution {
     func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
         var l = 0
         var r = nums.count - 1
-        
+
         while l < r {
             let sum = nums[l] + nums[r]
             if sum == target {
@@ -230,7 +230,7 @@ class Solution {
                 l += 1
             }
         }
-        
+
         return []
     }
 }

lcof/面试题58 - I. 翻转单词顺序/README.md

@@ -258,7 +258,7 @@ class Solution {
     func reverseWords(_ s: String) -> String {
         var words = [String]()
         var i = s.startIndex
-        
+
         while i < s.endIndex {
             while i < s.endIndex && s[i] == " " {
                 i = s.index(after: i)
@@ -274,7 +274,7 @@ class Solution {
                 i = j
             }
         }
-        
+
         words.reverse()
         return words.joined(separator: " ")
     }

lcof/面试题59 - I. 滑动窗口的最大值/README.md

@@ -262,23 +262,23 @@ class Solution {
         let n = nums.count
         var ans = [Int]()
         var deque = [Int]()
-        
+
         for i in 0..<n {
             if !deque.isEmpty && deque.first! < i - k + 1 {
                 deque.removeFirst()
             }
-            
+
             while !deque.isEmpty && nums[deque.last!] <= nums[i] {
                 deque.removeLast()
             }
-            
+
             deque.append(i)
-            
+
             if i >= k - 1 {
                 ans.append(nums[deque.first!])
             }
         }
-        
+
         return ans
     }
 }

lcof/面试题59 - II. 队列的最大值/README.md

@@ -428,22 +428,22 @@ public class MaxQueue {
 class MaxQueue {
     private var q1: [Int] = []
     private var q2: [Int] = []
-    
+
     init() {
     }
-    
+
     func max_value() -> Int {
         return q2.isEmpty ? -1 : q2.first!
     }
-    
+
     func push_back(_ value: Int) {
         q1.append(value)
         while !q2.isEmpty && q2.last! < value {
             q2.removeLast()
         }
         q2.append(value)
     }
-    
+
     func pop_front() -> Int {
         if q1.isEmpty {
             return -1

solution/3100-3199/3167.Better Compression of String/README.md

@@ -0,0 +1,220 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3167.Better%20Compression%20of%20String/README.md
+---
+
+<!-- problem:start -->
+
+# [3167. Better Compression of String 🔒](https://leetcode.cn/problems/better-compression-of-string)
+
+[English Version](/solution/3100-3199/3167.Better%20Compression%20of%20String/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>You are given a string <code>compressed</code> representing a compressed version of a string. The format is a character followed by its frequency. For example, <code>&quot;a3b1a1c2&quot;</code> is a compressed version of the string <code>&quot;aaabacc&quot;</code>.</p>
+
+<p>We seek a <strong>better compression</strong> with the following conditions:</p>
+
+<ol>
+	<li>Each character should appear <strong>only once</strong> in the compressed version.</li>
+	<li>The characters should be in <strong>alphabetical order</strong>.</li>
+</ol>
+
+<p>Return the <em>better compression</em> of <code>compressed</code>.</p>
+
+<p><strong>Note:</strong> In the better version of compression, the order of letters may change, which is acceptable.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">compressed = &quot;a3c9b2c1&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;a3b2c10&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Characters &quot;a&quot; and &quot;b&quot; appear only once in the input, but &quot;c&quot; appears twice, once with a size of 9 and once with a size of 1.</p>
+
+<p>Hence, in the resulting string, it should have a size of 10.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">compressed = &quot;c2b3a1&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;a1b3c2&quot;</span></p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">compressed = &quot;a2b4c1&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;a2b4c1&quot;</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= compressed.length &lt;= 6 * 10<sup>4</sup></code></li>
+	<li><code>compressed</code> consists only of lowercase English letters and digits.</li>
+	<li><code>compressed</code> is a valid compression, i.e., each character is followed by its frequency.</li>
+	<li>Frequencies are in the range <code>[1, 10<sup>4</sup>]</code> and have no leading zeroes.</li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：哈希表 + 双指针
+
+我们可以使用哈希表来统计每个字符的频率，然后使用双指针来遍历 `compressed` 字符串，将每个字符的频率累加到哈希表中，最后按照字母顺序将字符和频率拼接成字符串。
+
+时间复杂度 $O(n + |\Sigma| \log |\Sigma|)$，空间复杂度 $O(|\Sigma|)$，其中 $n$ 是字符串 `compressed` 的长度，而 $|\Sigma|$ 是字符集的大小，这里字符集是小写字母，所以 $|\Sigma| = 26$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def betterCompression(self, compressed: str) -> str:
+        cnt = Counter()
+        i, n = 0, len(compressed)
+        while i < n:
+            j = i + 1
+            x = 0
+            while j < n and compressed[j].isdigit():
+                x = x * 10 + int(compressed[j])
+                j += 1
+            cnt[compressed[i]] += x
+            i = j
+        return "".join(sorted(f"{k}{v}" for k, v in cnt.items()))
+```
+
+#### Java
+
+```java
+class Solution {
+    public String betterCompression(String compressed) {
+        Map<Character, Integer> cnt = new TreeMap<>();
+        int i = 0;
+        int n = compressed.length();
+        while (i < n) {
+            char c = compressed.charAt(i);
+            int j = i + 1;
+            int x = 0;
+            while (j < n && Character.isDigit(compressed.charAt(j))) {
+                x = x * 10 + (compressed.charAt(j) - '0');
+                j++;
+            }
+            cnt.merge(c, x, Integer::sum);
+            i = j;
+        }
+        StringBuilder ans = new StringBuilder();
+        for (var e : cnt.entrySet()) {
+            ans.append(e.getKey()).append(e.getValue());
+        }
+        return ans.toString();
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    string betterCompression(string compressed) {
+        map<char, int> cnt;
+        int i = 0;
+        int n = compressed.length();
+        while (i < n) {
+            char c = compressed[i];
+            int j = i + 1;
+            int x = 0;
+            while (j < n && isdigit(compressed[j])) {
+                x = x * 10 + (compressed[j] - '0');
+                j++;
+            }
+            cnt[c] += x;
+            i = j;
+        }
+        stringstream ans;
+        for (const auto& entry : cnt) {
+            ans << entry.first << entry.second;
+        }
+        return ans.str();
+    }
+};
+```
+
+#### Go
+
+```go
+func betterCompression(compressed string) string {
+	cnt := map[byte]int{}
+	n := len(compressed)
+	for i := 0; i < n; {
+		c := compressed[i]
+		j := i + 1
+		x := 0
+		for j < n && compressed[j] >= '0' && compressed[j] <= '9' {
+			x = x*10 + int(compressed[j]-'0')
+			j++
+		}
+		cnt[c] += x
+		i = j
+	}
+	ans := strings.Builder{}
+	for c := byte('a'); c <= byte('z'); c++ {
+		if cnt[c] > 0 {
+			ans.WriteByte(c)
+			ans.WriteString(strconv.Itoa(cnt[c]))
+		}
+	}
+	return ans.String()
+}
+```
+
+#### TypeScript
+
+```ts
+function betterCompression(compressed: string): string {
+    const cnt = new Map<string, number>();
+    const n = compressed.length;
+    let i = 0;
+
+    while (i < n) {
+        const c = compressed[i];
+        let j = i + 1;
+        let x = 0;
+        while (j < n && /\d/.test(compressed[j])) {
+            x = x * 10 + +compressed[j];
+            j++;
+        }
+        cnt.set(c, (cnt.get(c) || 0) + x);
+        i = j;
+    }
+    const keys = Array.from(cnt.keys()).sort();
+    const ans: string[] = [];
+    for (const k of keys) {
+        ans.push(`${k}${cnt.get(k)}`);
+    }
+    return ans.join('');
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->