refactor 727

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_727.java

@@ -2,27 +2,6 @@
 
 import java.util.Arrays;
 
-/**
- * 727. Minimum Window Subsequence
- *
- * Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequence of W.
- * If there is no such window in S that covers all characters in T,
- * return the empty string "". If there are multiple such minimum-length windows, return the one with the left-most starting index.
-
- Example 1:
- Input:
- S = "abcdebdde", T = "bde"
- Output: "bcde"
-
- Explanation:
- "bcde" is the answer because it occurs before "bdde" which has the same length.
- "deb" is not a smaller window because the elements of T in the window must occur in order.
-
- Note:
- All the strings in the input will only contain lowercase letters.
- The length of S will be in the range [1, 20000].
- The length of T will be in the range [1, 100].
- */
 public class _727 {
     public static class Solution1 {
         /**
@@ -53,7 +32,9 @@ private boolean isSubsequence(String T, String sub) {
     }
 
     public static class Solution2 {
-        /**credit: https://github.com/lydxlx1/LeetCode/blob/master/src/_727.java*/
+        /**
+         * credit: https://github.com/lydxlx1/LeetCode/blob/master/src/_727.java
+         */
         public String minWindow(String S, String T) {
             int[][] dp = new int[S.length() + 1][T.length() + 1];
             int INFINITY = 1 << 29;