refactor 161

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_161.java

@@ -6,12 +6,10 @@
  * Given two strings s and t, determine if they are both one edit distance apart.
  *
  * Note:
- *
  * There are 3 possiblities to satisify one edit distance apart:
- *
- *     Insert a character into s to get t
- *     Delete a character from s to get t
- *     Replace a character of s to get t
+ * Insert a character into s to get t
+ * Delete a character from s to get t
+ * Replace a character of s to get t
  *
  * Example 1:
  * Input: s = "ab", t = "acb"
@@ -29,43 +27,43 @@
  * Explanation: We can replace '0' with '1' to get t.
  */
 public class _161 {
-  public static class Solution1 {
-    public boolean isOneEditDistance(String s, String t) {
-      char[] schar = s.toCharArray();
-      char[] tchar = t.toCharArray();
+    public static class Solution1 {
+        public boolean isOneEditDistance(String s, String t) {
+            char[] schar = s.toCharArray();
+            char[] tchar = t.toCharArray();
 
-      if (Math.abs(s.length() - t.length()) == 1) {
-        char[] longer = (s.length() > t.length()) ? schar : tchar;
-        char[] shorter = (longer == schar) ? tchar : schar;
+            if (Math.abs(s.length() - t.length()) == 1) {
+                char[] longer = (s.length() > t.length()) ? schar : tchar;
+                char[] shorter = (longer == schar) ? tchar : schar;
 
-        int diffCnt = 0;
-        int i = 0;
-        int j = 0;
-        for (; i < shorter.length && j < longer.length; ) {
-          if (longer[j] != shorter[i]) {
-            diffCnt++;
-            j++;
-          } else {
-            i++;
-            j++;
-          }
-        }
-        return diffCnt == 1
-            || diffCnt
-            == 0;//it could be the last char of the longer is the different one, in that case, diffCnt remains to be zero
-      } else if (s.length() == t.length()) {
-        int diffCnt = 0;
-        for (int i = 0; i < s.length(); i++) {
-          if (schar[i] != tchar[i]) {
-            diffCnt++;
-          }
-          if (diffCnt > 1) {
+                int diffCnt = 0;
+                int i = 0;
+                int j = 0;
+                for (; i < shorter.length && j < longer.length; ) {
+                    if (longer[j] != shorter[i]) {
+                        diffCnt++;
+                        j++;
+                    } else {
+                        i++;
+                        j++;
+                    }
+                }
+                return diffCnt == 1
+                        || diffCnt
+                        == 0;//it could be the last char of the longer is the different one, in that case, diffCnt remains to be zero
+            } else if (s.length() == t.length()) {
+                int diffCnt = 0;
+                for (int i = 0; i < s.length(); i++) {
+                    if (schar[i] != tchar[i]) {
+                        diffCnt++;
+                    }
+                    if (diffCnt > 1) {
+                        return false;
+                    }
+                }
+                return diffCnt == 1;
+            }
             return false;
-          }
         }
-        return diffCnt == 1;
-      }
-      return false;
     }
-  }
 }