refactor 760

fishercoder1534 · fishercoder1534 · commit e386b18683e9 · 2020-07-22T15:59:06.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_760.java b/src/main/java/com/fishercoder/solutions/_760.java
@@ -1,38 +1,17 @@
 package com.fishercoder.solutions;
 
-/**
- * 760. Find Anagram Mappings
- *
- * Given two lists Aand B, and B is an anagram of A. B is an anagram of A means B is made by randomizing the order of the elements in A.
- * We want to find an index mapping P, from A to B. A mapping P[i] = j means the ith element in A appears in B at index j.
- * These lists A and B may contain duplicates. If there are multiple answers, output any of them.
-
- For example, given
-
- A = [12, 28, 46, 32, 50]
- B = [50, 12, 32, 46, 28]
-
- We should return
- [1, 4, 3, 2, 0]
- as P[0] = 1 because the 0th element of A appears at B[1], and P[1] = 4 because the 1st element of A appears at B[4], and so on.
-
- Note:
-
- A, B have equal lengths in range [1, 100].
- A[i], B[i] are integers in range [0, 10^5].
- */
 public class _760 {
-  public static class Solution1 {
-    public int[] anagramMappings(int[] A, int[] B) {
-      int[] result = new int[A.length];
-      for (int i = 0; i < A.length; i++) {
-        for (int j = 0; j < B.length; j++) {
-          if (A[i] == B[j]) {
-            result[i] = j;
-          }
+    public static class Solution1 {
+        public int[] anagramMappings(int[] A, int[] B) {
+            int[] result = new int[A.length];
+            for (int i = 0; i < A.length; i++) {
+                for (int j = 0; j < B.length; j++) {
+                    if (A[i] == B[j]) {
+                        result[i] = j;
+                    }
+                }
+            }
+            return result;
         }
-      }
-      return result;
     }
-  }
 }
