[N-0] refactor 567

Author: fishercoder1534

README.md

@@ -123,7 +123,7 @@ Your ideas/fixes/algorithms are more than welcome!
 |573|[Squirrel Simulation](https://leetcode.com/problems/squirrel-simulation/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_573.java) | O(n) |O(1) | Medium | Math
 |572|[Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_572.java) | O(m*n) |O(1) | Easy | Tree
 |568|[Maximum Vacation Days](https://leetcode.com/problems/maximum-vacation-days/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_568.java) | O(n^2*k) |O(n*k) | Hard | DP
-|567|[Permutation in String](https://leetcode.com/problems/permutation-in-string/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_567.java) | O(max(m,n)) |O(1) | Medium | Sliding Windows, Two Pointers
+|567|[Permutation in String](https://leetcode.com/problems/permutation-in-string/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_567.java) | O(l1 + 26*(l2 - l1)) |O(1) | Medium | Sliding Windows, Two Pointers
 |566|[Reshape the Matrix](https://leetcode.com/problems/reshape-the-matrix/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_566.java) | O(m*n) |O(1) | Easy | 
 |565|[Array Nesting](https://leetcode.com/problems/array-nesting/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_565.java) | O(n) |O(n) | Medium | 
 |563|[Binary Tree Tilt](https://leetcode.com/problems/binary-tree-tilt/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_563.java) | O(n) |O(n) | Easy | Tree Recursion

src/main/java/com/fishercoder/solutions/_567.java

@@ -21,57 +21,45 @@
  */
 public class _567 {
 
-    //credit: sliding window: https://discuss.leetcode.com/topic/87845/java-solution-sliding-window
-    /**1. How do we know string p is a permutation of string s? Easy, each character in p is in s too.
-     * So we can abstract all permutation strings of s to a map (Character -> Count). i.e. abba -> {a:2, b:2}.
-     * Since there are only 26 lower case letters in this problem, we can just use an array to represent the map.
-
-     2. How do we know string s2 contains a permutation of s1?
-     We just need to create a sliding window with length of s1,
-     move from beginning to the end of s2.
-     When a character moves in from right of the window,
-     we subtract 1 to that character count from the map.
-     When a character moves out from left of the window,
-     we add 1 to that character count. So once we see all zeros in the map,
-     meaning equal numbers of every characters between s1 and the substring in the sliding window, we know the answer is true.
-     */
-    public boolean checkInclusion(String s1, String s2) {
-        int len1 = s1.length();
-        int len2 = s2.length();
-        if (len1 > len2) {
-            return false;
-        }
-
-        int[] count = new int[26];
-        for (int i = 0; i < len1; i++) {
-            count[s1.charAt(i) - 'a']++;
-        }
-
-        for (int i = 0; i < len1; i++) {
-            count[s2.charAt(i) - 'a']--;
-        }
+    public static class Solution1 {
+        /**
+         * credit: sliding window: https://discuss.leetcode.com/topic/87845/java-solution-sliding-window
+         */
+        public boolean checkInclusion(String s1, String s2) {
+            int len1 = s1.length();
+            int len2 = s2.length();
+            if (len1 > len2) {
+                return false;
+            }
 
-        if (allZeroes(count)) {
-            return true;
-        }
+            int[] count = new int[26];
+            for (int i = 0; i < len1; i++) {
+                count[s1.charAt(i) - 'a']++;
+                count[s2.charAt(i) - 'a']--;
+            }
 
-        for (int i = len1; i < len2; i++) {
-            count[s2.charAt(i) - 'a']--;
-            count[s2.charAt(i - len1) - 'a']++;
             if (allZeroes(count)) {
                 return true;
             }
-        }
 
-        return false;
-    }
+            for (int i = len1; i < len2; i++) {
+                count[s2.charAt(i) - 'a']--;
+                count[s2.charAt(i - len1) - 'a']++;
+                if (allZeroes(count)) {
+                    return true;
+                }
+            }
 
-    private boolean allZeroes(int[] count) {
-        for (int i : count) {
-            if (i != 0) {
-                return false;
+            return false;
+        }
+
+        private boolean allZeroes(int[] count) {
+            for (int i : count) {
+                if (i != 0) {
+                    return false;
+                }
             }
+            return true;
         }
-        return true;
     }
 }

src/test/java/com/fishercoder/_567Test.java

@@ -6,9 +6,6 @@
 
 import static org.junit.Assert.assertEquals;
 
-/**
- * Created by fishercoder on 4/30/17.
- */
 public class _567Test {
     private static _567 test;
     private static boolean expected;