refactor 868

fishercoder1534 · fishercoder1534 · commit 5f360e2216d3 · 2020-07-25T06:42:38.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_868.java b/src/main/java/com/fishercoder/solutions/_868.java
@@ -3,60 +3,21 @@
 import java.util.ArrayList;
 import java.util.List;
 
-/**
- * 868. Binary Gap
- *
- * Given a positive integer N, find and return the longest distance between two consecutive 1's in the binary representation of N.
- *
- * If there aren't two consecutive 1's, return 0.
- *
- * Example 1:
- * Input: 22
- * Output: 2
- * Explanation:
- * 22 in binary is 0b10110.
- * In the binary representation of 22, there are three ones, and two consecutive pairs of 1's.
- * The first consecutive pair of 1's have distance 2.
- * The second consecutive pair of 1's have distance 1.
- * The answer is the largest of these two distances, which is 2.
- *
- * Example 2:
- * Input: 5
- * Output: 2
- * Explanation:
- * 5 in binary is 0b101.
- *
- * Example 3:
- * Input: 6
- * Output: 1
- * Explanation:
- * 6 in binary is 0b110.
- *
- * Example 4:
- * Input: 8
- * Output: 0
- * Explanation:
- * 8 in binary is 0b1000.
- * There aren't any consecutive pairs of 1's in the binary representation of 8, so we return 0.
- *
- * Note:
- * 1 <= N <= 10^9
- */
 public class _868 {
-  public static class Solution1 {
-    public int binaryGap(int N) {
-      String bin = Integer.toBinaryString(N);
-      List<Integer> oneIndexes = new ArrayList<>();
-      for (int i = 0; i < bin.length(); i++) {
-        if (bin.charAt(i) == '1') {
-          oneIndexes.add(i);
+    public static class Solution1 {
+        public int binaryGap(int N) {
+            String bin = Integer.toBinaryString(N);
+            List<Integer> oneIndexes = new ArrayList<>();
+            for (int i = 0; i < bin.length(); i++) {
+                if (bin.charAt(i) == '1') {
+                    oneIndexes.add(i);
+                }
+            }
+            int maxGap = 0;
+            for (int i = 0; i < oneIndexes.size() - 1; i++) {
+                maxGap = Math.max(oneIndexes.get(i + 1) - oneIndexes.get(i), maxGap);
+            }
+            return maxGap;
         }
-      }
-      int maxGap = 0;
-      for (int i = 0; i < oneIndexes.size() - 1; i++) {
-        maxGap = Math.max(oneIndexes.get(i + 1) - oneIndexes.get(i), maxGap);
-      }
-      return maxGap;
     }
-  }
 }
