refactor 115

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_115.java

@@ -1,44 +1,33 @@
 package com.fishercoder.solutions;
 
-/**
- * 115. Distinct Subsequences
- *
- * Given a string S and a string T, count the number of distinct subsequences of S which equals T. A
- * subsequence of a string is a new string which is formed from the original string by deleting some
- * (can be none) of the characters without disturbing the relative positions of the remaining
- * characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
- *
- * Here is an example: S = "rabbbit", T = "rabbit" Return 3.
- */
-
 public class _115 {
-  public static class Solution1 {
-    public int numDistinct(String s, String t) {
-      int m = s.length();
-      int n = t.length();
-      int[][] dp = new int[m + 1][n + 1];
+    public static class Solution1 {
+        public int numDistinct(String s, String t) {
+            int m = s.length();
+            int n = t.length();
+            int[][] dp = new int[m + 1][n + 1];
 
-      char[] schar = s.toCharArray();
-      char[] tchar = t.toCharArray();
+            char[] schar = s.toCharArray();
+            char[] tchar = t.toCharArray();
 
-      for (int i = 0; i <= m; i++) {
-        dp[i][0] = 1;
-      }
+            for (int i = 0; i <= m; i++) {
+                dp[i][0] = 1;
+            }
 
-      for (int j = 1; j <= n; j++) {
-        dp[0][j] = 0;
-      }
+            for (int j = 1; j <= n; j++) {
+                dp[0][j] = 0;
+            }
 
-      for (int i = 1; i <= m; i++) {
-        for (int j = 1; j <= n; j++) {
-          if (schar[i - 1] == tchar[j - 1]) {
-            dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
-          } else {
-            dp[i][j] = dp[i - 1][j];
-          }
+            for (int i = 1; i <= m; i++) {
+                for (int j = 1; j <= n; j++) {
+                    if (schar[i - 1] == tchar[j - 1]) {
+                        dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
+                    } else {
+                        dp[i][j] = dp[i - 1][j];
+                    }
+                }
+            }
+            return dp[m][n];
         }
-      }
-      return dp[m][n];
     }
-  }
 }