338. Counting Bits

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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

   1 public class Solution {
   2     public int[] countBits(int num) {
   3         int[] result = new int[num+1];
   4         for(int i=0;i<num+1;i++){
   5             int count = 0;
   6             int j = i;
   7             while(j!=0){
   8                 j=j&(j-1);
   9                 count++;
  10             }
  11             result[i] = count;
  12         }
  13         return result;
  14     }
  15 }

  public int[] countBits(int num) {for (int n = 0; n <= num; ++n) {
              // binaryNum接收转为二进制的数字
              String binaryNum = Integer.toBinaryString(n);
              char[] charArr = binaryNum.toCharArray();
              int i = 0;
              for (int m = 0; m < charArr.length; ++m) {
                  if (charArr[m] == ‘1‘) {
                      i++;
                  }
              }
              res[n] = i;
          }return res;
      }

  第二个解法是网上看到，第一个解法根据编程之美书中求一个数中二进制表示中的 1 的位数得来。

338. Counting Bits

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原文地址：http://www.cnblogs.com/guoguolan/p/5390467.html