refactor 76

fishercoder1534 · fishercoder1534 · commit 1d803cb42757 · 2020-05-05T07:09:45.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_76.java b/src/main/java/com/fishercoder/solutions/_76.java
@@ -1,61 +1,44 @@
 package com.fishercoder.solutions;
 
-/**
- * 76. Minimum Window Substring
- *
- * Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
-
- For example,
- S = "ADOBECODEBANC"
- T = "ABC"
-
- Minimum window is "BANC".
-
- Note:
-
- If there is no such window in S that covers all characters in T, return the empty string "".
- If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
- */
-
 public class _76 {
 
-  public static class Solution1 {
-    public String minWindow(String s, String t) {
-      int[] counts = new int[256];
-      for (char c : t.toCharArray()) {
-        counts[c]++;
-      }
-
-      int start = 0;
-      int end = 0;
-      int minStart = 0;
-      int minLen = Integer.MAX_VALUE;
-      int counter = t.length();
-      while (end < s.length()) {
-        if (counts[s.charAt(end)] > 0) {
-          counter--;
+    public static class Solution1 {
+        public String minWindow(String s, String t) {
+            int[] counts = new int[256];
+            for (char c : t.toCharArray()) {
+                counts[c]++;
+            }
+
+            int start = 0;
+            int end = 0;
+            int minStart = 0;
+            int minLen = Integer.MAX_VALUE;
+            int counter = t.length();
+            while (end < s.length()) {
+                if (counts[s.charAt(end)] > 0) {
+                    counter--;
+                }
+
+                counts[s.charAt(end)]--;
+                end++;
+
+                while (counter == 0) {
+                    if (end - start < minLen) {
+                        minStart = start;
+                        minLen = end - start;
+                    }
+                    counts[s.charAt(start)]++;
+                    if (counts[s.charAt(start)] > 0) {
+                        counter++;
+                    }
+                    start++;
+                }
+            }
+
+            if (minLen == Integer.MAX_VALUE) {
+                return "";
+            }
+            return s.substring(minStart, minStart + minLen);
         }
-
-        counts[s.charAt(end)]--;
-        end++;
-
-        while (counter == 0) {
-          if (end - start < minLen) {
-            minStart = start;
-            minLen = end - start;
-          }
-          counts[s.charAt(start)]++;
-          if (counts[s.charAt(start)] > 0) {
-            counter++;
-          }
-          start++;
-        }
-      }
-
-      if (minLen == Integer.MAX_VALUE) {
-        return "";
-      }
-      return s.substring(minStart, minStart + minLen);
     }
-  }
 }
