refactor 338

fishercoder1534 · fishercoder1534 · commit 37c1a84d0f45 · 2020-05-28T06:09:47.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_338.java b/src/main/java/com/fishercoder/solutions/_338.java
@@ -1,26 +1,5 @@
 package com.fishercoder.solutions;
 
-/**338. Counting Bits
-Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.
-
-Example:
-For num = 5 you should return [0,1,1,2,1,2].
-
-Follow up:
-
-It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
-Space complexity should be O(n).
-Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.
-
-Hint:
-
-You should make use of what you have produced already.
-Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
-Or does the odd/even status of the number help you in calculating the number of 1s?
-
-*
-*
-*/
 public class _338 {
     public static class Solution1 {
         //use the most regular method to get it AC'ed first
@@ -43,12 +22,14 @@ private int countOnes(int i) {
     }
 
     private class Solution2 {
-        /**lixx2100's post is cool:https://discuss.leetcode.com/topic/40162/three-line-java-solution
-        An easy recurrence for this problem is f[i] = f[i / 2] + i % 2
-        and then we'll use bit manipulation to express the above recursion function
-         right shift by 1 means to divide by 2
-        AND with 1 means to modulo 2
-        this is so cool!*/
+        /**
+         * lixx2100's post is cool:https://discuss.leetcode.com/topic/40162/three-line-java-solution
+         * An easy recurrence for this problem is f[i] = f[i / 2] + i % 2
+         * and then we'll use bit manipulation to express the above recursion function
+         * right shift by 1 means to divide by 2
+         * AND with 1 means to modulo 2
+         * this is so cool!
+         */
         public int[] countBits(int num) {
             int[] ones = new int[num + 1];
             for (int i = 1; i <= num; i++) {
