edit 438

Author: fishercoder1534

README.md

@@ -183,7 +183,7 @@ Your ideas/fixes/algorithms are more than welcome!
 |441|[Arranging Coins](https://leetcode.com/problems/arrange-coins/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_441.java)| O(n)|O(1) | Easy|
 |440|[K-th Smallest in Lexicographical Order](https://leetcode.com/problems/k-th-smallest-in-lexicographical-order/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_440.java)| O(n^2)|O(1) | Hard|
 |439|[Ternary Expression Parser](https://leetcode.com/problems/ternary-expression-parser/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_439.java)| O(n)|O(n) | Medium| Stack
-|438|[Find All Anagrams in a String](https://leetcode.com/problems/find-all-anagrams-in-a-string/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_438.java)| O(n)|O(1) | Easy|
+|438|[Find All Anagrams in a String](https://leetcode.com/problems/find-all-anagrams-in-a-string/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_438.java)| O(n)|O(1) | Easy| Sliding Window
 |437|[Path Sum III](https://leetcode.com/problems/path-sum-iii/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_437.java) | O(n^2) |O(n) | Easy| DFS, recursion
 |436|[Find Right Interval](https://leetcode.com/problems/find-right-interval/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_436.java) | O(nlogn) |O(n) | Medium| Binary Search
 |435|[Non-overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_435.java) | O(nlogn) |O(1) | Medium| Greedy

src/main/java/com/fishercoder/solutions/_438.java

@@ -3,59 +3,87 @@
 import java.util.ArrayList;
 import java.util.List;
 
+/**438. Find All Anagrams in a String
+ * Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
+ Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
+
+ The order of output does not matter.
+
+ Example 1:
+ Input:
+ s: "cbaebabacd" p: "abc"
+ Output:
+ [0, 6]
+
+ Explanation:
+ The substring with start index = 0 is "cba", which is an anagram of "abc".
+ The substring with start index = 6 is "bac", which is an anagram of "abc".
+
+ Example 2:
+ Input:
+ s: "abab" p: "ab"
+ Output:
+ [0, 1, 2]
+
+ Explanation:
+ The substring with start index = 0 is "ab", which is an anagram of "ab".
+ The substring with start index = 1 is "ba", which is an anagram of "ab".
+ The substring with start index = 2 is "ab", which is an anagram of "ab".*/
+
 public class _438 {
-    /**O(m*n) solution, my original and most intuitive one, but kind of brute force.*/
+    /**
+     * O(m*n) solution, my original and most intuitive one, but kind of brute force.
+     */
     public List<Integer> findAnagrams(String s, String p) {
         List<Integer> result = new ArrayList();
-        for (int i = 0; i <= s.length()-p.length(); i++){
-            if (isAnagram(s.substring(i, i+p.length()), p)) result.add(i);
+        for (int i = 0; i <= s.length() - p.length(); i++) {
+            if (isAnagram(s.substring(i, i + p.length()), p)) result.add(i);
         }
         return result;
     }
-    
-    private boolean isAnagram(String s, String p){
+
+    private boolean isAnagram(String s, String p) {
         int[] c = new int[26];
-        for (int i = 0; i < s.length(); i++){
+        for (int i = 0; i < s.length(); i++) {
             c[s.charAt(i) - 'a']++;
             c[p.charAt(i) - 'a']--;
         }
-        
-        for (int i : c){
-            if(i != 0) return false;
+
+        for (int i : c) {
+            if (i != 0) return false;
         }
         return true;
     }
 
-    
+
     static class SlidingWindowSolution {
-        /**O(n) solution inspired by this post: https://discuss.leetcode.com/topic/64434/shortest-concise-java-o-n-sliding-window-solution*/
         public List<Integer> findAnagrams(String s, String p) {
             List<Integer> result = new ArrayList();
             int[] hash = new int[26];
-            for (char c : p.toCharArray()){
+            for (char c : p.toCharArray()) {
                 hash[c - 'a']++;
             }
             int start = 0, end = 0, count = p.length();
-            while (end < s.length()){
-                if(hash[s.charAt(end) - 'a'] > 0){
+            while (end < s.length()) {
+                if (hash[s.charAt(end) - 'a'] > 0) {
                     count--;
                 }
                 hash[s.charAt(end) - 'a']--;
                 end++;
-                
-                if(count == 0) result.add(start);
-                
-                if((end - start) == p.length()){
-                    if(hash[s.charAt(start) - 'a'] >= 0) count++;
+
+                if (count == 0) result.add(start);
+
+                if ((end - start) == p.length()) {
+                    if (hash[s.charAt(start) - 'a'] >= 0) count++;
                     hash[s.charAt(start) - 'a']++;
                     start++;
                 }
             }
             return result;
         }
     }
-    
-    public static void main(String...args){
+
+    public static void main(String... args) {
         SlidingWindowSolution test = new SlidingWindowSolution();
         String s = "cbaebabacd";
         String p = "abc";