refactor 467

fishercoder1534 · fishercoder1534 · commit 64944daad7a7 · 2019-04-27T07:44:41.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_467.java b/src/main/java/com/fishercoder/solutions/_467.java
@@ -1,61 +1,65 @@
 package com.fishercoder.solutions;
 
-/**Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".
-
- Now we have another string p. Your job is to find out how many unique non-empty substrings of p are present in s. In particular, your input is the string p and you need to output the number of different non-empty substrings of p in the string s.
-
- Note: p consists of only lowercase English letters and the size of p might be over 10000.
+/**
+ * 467. Unique Substrings in Wraparound String
+ *
+ * Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz",
+ * so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".
+ * Now we have another string p.
+ * Your job is to find out how many unique non-empty substrings of p are present in s.
+ * In particular, your input is the string p and you need to output the number of different non-empty substrings of p in the string s.
+ * Note: p consists of only lowercase English letters and the size of p might be over 10000.
 
  Example 1:
  Input: "a"
  Output: 1
-
  Explanation: Only the substring "a" of string "a" is in the string s.
+
  Example 2:
  Input: "cac"
  Output: 2
  Explanation: There are two substrings "a", "c" of string "cac" in the string s.
+
  Example 3:
  Input: "zab"
  Output: 6
- Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.*/
+ Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.
+ */
 public class _467 {
-    /**A naive solution would definitely result in TLE.
-     * Since the pattern keeps repeating, DP is the way to go.
-     * Because the input consists merely of lowercase English letters, we could maintain an array of size 26,
-     * keep updating this array, counting the substrings that end with this letter, keep updating it with the largest one possible.
-     * 
-     * Inspired by this: https://discuss.leetcode.com/topic/70658/concise-java-solution-using-dp*/
-    public static int findSubstringInWraproundString(String p) {
-        if (p == null || p.isEmpty()) {
-            return 0;
-        }
-        if (p.length() < 2) {
-            return p.length();
-        }
-        int count = 0;
-        int[] dp = new int[26];
-        dp[p.charAt(0) - 'a'] = 1;
-        int len = 1;
-        for (int i = 1; i < p.length(); i++) {
-            if (p.charAt(i) - 1 == p.charAt(i - 1) || (p.charAt(i) == 'a' && p.charAt(i - 1) == 'z')) {
-                len++;
-            } else {
-                len = 1;
+    public static class Solution1 {
+        /**
+         * A naive solution would definitely result in TLE.
+         * Since the pattern keeps repeating, DP is the way to go.
+         * Because the input consists merely of lowercase English letters, we could maintain an array of size 26,
+         * keep updating this array, counting the substrings that end with this letter, keep updating it with the largest one possible.
+         * <p>
+         * Inspired by this: https://discuss.leetcode.com/topic/70658/concise-java-solution-using-dp
+         */
+        public int findSubstringInWraproundString(String p) {
+            if (p == null || p.isEmpty()) {
+                return 0;
+            }
+            if (p.length() < 2) {
+                return p.length();
+            }
+            int count = 0;
+            int[] dp = new int[26];
+            dp[p.charAt(0) - 'a'] = 1;
+            int len = 1;
+            for (int i = 1; i < p.length(); i++) {
+                if (p.charAt(i) - 1 == p.charAt(i - 1) || (p.charAt(i) == 'a' && p.charAt(i - 1) == 'z')) {
+                    len++;
+                } else {
+                    len = 1;
+                }
+                dp[p.charAt(i) - 'a'] = Math.max(len, dp[p.charAt(i) - 'a']);
             }
-            dp[p.charAt(i) - 'a'] = Math.max(len, dp[p.charAt(i) - 'a']);
-        }
 
-        for (int i : dp) {
-            count += i;
+            for (int i : dp) {
+                count += i;
+            }
+            return count;
         }
-        return count;
     }
 
-    public static void main(String... args) {
-//        String p = "a";
-//        String p = "abcgha";
-        String p = "zab";
-        System.out.println(findSubstringInWraproundString(p));
-    }
 }
diff --git a/src/test/java/com/fishercoder/_467Test.java b/src/test/java/com/fishercoder/_467Test.java
@@ -0,0 +1,23 @@
+package com.fishercoder;
+
+import com.fishercoder.solutions._467;
+import org.junit.BeforeClass;
+import org.junit.Test;
+
+import static junit.framework.Assert.assertEquals;
+
+public class _467Test {
+    private static _467.Solution1 solution1;
+
+    @BeforeClass
+    public static void setup() {
+        solution1 = new _467.Solution1();
+    }
+
+    @Test
+    public void test1() {
+        assertEquals(1, solution1.findSubstringInWraproundString("a"));
+
+    }
+
+}
