refactor 190

fishercoder1534 · fishercoder1534 · commit acaee0f44ffb · 2020-07-12T06:18:25.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_190.java b/src/main/java/com/fishercoder/solutions/_190.java
@@ -1,20 +1,5 @@
 package com.fishercoder.solutions;
 
-/**
- * 190. Reverse Bits
- * Reverse bits of a given 32 bits unsigned integer.
-
- Example:
- Input: 43261596
- Output: 964176192
-
- Explanation: 43261596 represented in binary as 00000010100101000001111010011100,
- return 964176192 represented in binary as 00111001011110000010100101000000.
-
- Follow up:
- If this function is called many times, how would you optimize it?
- */
-
 public class _190 {
     /**delimiting the binary string into 4 bits array will make it easier to see/visualize:
      * original binary format:
@@ -23,28 +8,30 @@ public class _190 {
      * 0011,1001,0111,1000,0010,1001,0100,0000
      * The most right side digit shifted to the most left side, the 2nd right side digit shifted to the 2nd left side, so forth..*/
 
-    /**This post: http://stackoverflow.com/questions/2811319/difference-between-and
+    /**
+     * This post: http://stackoverflow.com/questions/2811319/difference-between-and
      * gives a good explanation between logical right shift: ">>>" and arithmetic right shift: ">>".
      * Basically, ">>" preserves the most left bit and treats it as the sign for this number,
      * e.g. -2 represented in 8 bits is 11111110, thus -2 >> 1 will become 11111111, i.e. -1
      * notice its sign bit (the most left one bit) is preserved
      * However, logical right shift ">>>" doesn't care about the first bit on the most left,
      * it simply shifts every bit to the right.
-     * e.g. -2 >>> 1 would become 1111111111111111111111111111111, i.e. 2147483647*/
+     * e.g. -2 >>> 1 would become 1111111111111111111111111111111, i.e. 2147483647
+     */
 
- public static class Solution1 {
-      // you need treat n as an unsigned value
-      public int reverseBits(int n) {
-        int res = 0;
-        for (int i = 0; i < 32; i++) {
-          res += n & 1;//get the most right bit each time
-          n >>>= 1;//do UN-signed right shift by 1 each time
-          if (i < 31) {
-            res <<=
-                1;//shift this number to the left by 1 each time, so that eventually, this number is reversed
-          }
+    public static class Solution1 {
+        // you need treat n as an unsigned value
+        public int reverseBits(int n) {
+            int res = 0;
+            for (int i = 0; i < 32; i++) {
+                res += n & 1;//get the most right bit each time
+                n >>>= 1;//do UN-signed right shift by 1 each time
+                if (i < 31) {
+                    res <<=
+                            1;//shift this number to the left by 1 each time, so that eventually, this number is reversed
+                }
+            }
+            return res;
         }
-        return res;
-      }
     }
 }
