feat: add solutions to leetcode problem: No.0438. Find All Anagrams in a String

Author: yanglbme

solution/0400-0499/0438.Find All Anagrams in a String/README.md

@@ -46,27 +46,97 @@ s: "abab" p: "ab"
 起始索引等于 2 的子串是 "ab", 它是 "ab" 的字母异位词。
 </pre>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+“双指针 + 滑动窗口”求解。定义滑动窗口的左右两个指针 left、right，right 一步步往右走遍历 s 字符串，当 right 指针遍历到的字符加入 t 后超过 counter 的字符数量时，将滑动窗口左侧字符不断弹出，也就是 left 指针不断右移，直到符合要求为止。
+
+若滑动窗口长度等于字符串 p 的长度时，这时的 s 子字符串就是 p 的异位词。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def findAnagrams(self, s: str, p: str) -> List[int]:
+        counter = collections.Counter(p)
+        res = []
+        left = right = 0
+        t = collections.Counter()
+        while right < len(s):
+            t[s[right]] += 1
+            while t[s[right]] > counter[s[right]]:
+                t[s[left]] -= 1
+                left += 1
+            if right - left == len(p) - 1:
+                res.append(left)
+            right += 1
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
+“暴力”求解。利用哈希表 counter 统计字符串 p 中每个字符出现的次数。然后遍历字符串 s，判断子串 `s[i, i + p)` 中每个字符出现的次数是否与 counter 相同。若是，则将当前下标 i 添加到结果列表中。时间复杂度 `O(s.length * p.length)`。
+
 ```java
+class Solution {
+    public List<Integer> findAnagrams(String s, String p) {
+        int[] counter = new int[26];
+        for (int i = 0; i < p.length(); ++i) {
+            ++counter[p.charAt(i) - 'a'];
+        }
+        List<Integer> res = new ArrayList<>();
+        for (int i = 0; i <= s.length() - p.length(); ++i) {
+            int[] t = Arrays.copyOf(counter, counter.length);
+            boolean find = true;
+            for (int j = i; j < i + p.length(); ++j) {
+                if (--t[s.charAt(j) - 'a'] < 0) {
+                    find = false;
+                    break;
+                }
+            }
+            if (find) {
+                res.add(i);
+            }
+        }
+        return res;
+    }
+}
+```
 
+“双指针 + 滑动窗口”求解。
+
+```java
+class Solution {
+    public List<Integer> findAnagrams(String s, String p) {
+        int[] counter = new int[26];
+        for (int i = 0; i < p.length(); ++i) {
+            ++counter[p.charAt(i) - 'a'];
+        }
+        List<Integer> res = new ArrayList<>();
+        int left = 0, right = 0;
+        int[] t = new int[26];
+        while (right < s.length()) {
+            int i = s.charAt(right) - 'a';
+            ++t[i];
+            while (t[i] > counter[i]) {
+                --t[s.charAt(left) - 'a'];
+                ++left;
+            }
+            if (right - left == p.length() - 1) {
+                res.add(left);
+            }
+            ++right;
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0400-0499/0438.Find All Anagrams in a String/README_EN.md

@@ -44,13 +44,50 @@ The substring with start index = 2 is "ab", which is an anagram of &qu
 ### **Python3**
 
 ```python
-
+class Solution:
+    def findAnagrams(self, s: str, p: str) -> List[int]:
+        counter = collections.Counter(p)
+        res = []
+        left = right = 0
+        t = collections.Counter()
+        while right < len(s):
+            t[s[right]] += 1
+            while t[s[right]] > counter[s[right]]:
+                t[s[left]] -= 1
+                left += 1
+            if right - left == len(p) - 1:
+                res.append(left)
+            right += 1
+        return res
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public List<Integer> findAnagrams(String s, String p) {
+        int[] counter = new int[26];
+        for (int i = 0; i < p.length(); ++i) {
+            ++counter[p.charAt(i) - 'a'];
+        }
+        List<Integer> res = new ArrayList<>();
+        int left = 0, right = 0;
+        int[] t = new int[26];
+        while (right < s.length()) {
+            int i = s.charAt(right) - 'a';
+            ++t[i];
+            while (t[i] > counter[i]) {
+                --t[s.charAt(left) - 'a'];
+                ++left;
+            }
+            if (right - left == p.length() - 1) {
+                res.add(left);
+            }
+            ++right;
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0400-0499/0438.Find All Anagrams in a String/Solution.java

@@ -0,0 +1,24 @@
+class Solution {
+    public List<Integer> findAnagrams(String s, String p) {
+        int[] counter = new int[26];
+        for (int i = 0; i < p.length(); ++i) {
+            ++counter[p.charAt(i) - 'a'];
+        }
+        List<Integer> res = new ArrayList<>();
+        int left = 0, right = 0;
+        int[] t = new int[26];
+        while (right < s.length()) {
+            int i = s.charAt(right) - 'a';
+            ++t[i];
+            while (t[i] > counter[i]) {
+                --t[s.charAt(left) - 'a'];
+                ++left;
+            }
+            if (right - left == p.length() - 1) {
+                res.add(left);
+            }
+            ++right;
+        }
+        return res;
+    }
+}

solution/0400-0499/0438.Find All Anagrams in a String/Solution.py

@@ -1,23 +1,15 @@
 class Solution:
-    def findAnagrams(self, s, p):
-        """
-        :type s: str
-        :type p: str
-        :rtype: List[int]
-        """
-        lens=len(s)
-        lenp=len(p)
-        if lens < lenp:
-            return []
-        flag1 = [0] * 26
-        flag2 = [0] * 26
-        for i, x in enumerate(p):
-            flag1[ord(x) - 97] += 1
-        ans = []
-        for i, x in enumerate(s):
-            flag2[ord(x)- 97] += 1
-            if i >= lenp:
-                flag2[ord(s[i - lenp]) - 97] -= 1
-            if flag1 == flag2:
-                ans.append(i-lenp+1)
-        return ans
+    def findAnagrams(self, s: str, p: str) -> List[int]:
+        counter = collections.Counter(p)
+        res = []
+        left = right = 0
+        t = collections.Counter()
+        while right < len(s):
+            t[s[right]] += 1
+            while t[s[right]] > counter[s[right]]:
+                t[s[left]] -= 1
+                left += 1
+            if right - left == len(p) - 1:
+                res.append(left)
+            right += 1
+        return res