[LeetCode]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.
eg:假如只有1位,则能表示的数字为0,1,相应的1出现的次数是0,1
当位数增加一位时,能表示的数字增加了2(10),3(11),相应的1出现的次数为1(0+1),2(1+1);
当位数增加到三位时,能表示的数字增加了4(100),5(101),6(110),7(111),1出现的次数为:0+1,1+1,1+1,2+1;
以此类推
public class Solution { public int[] countBits(int num) { int [] res = new int[num+1]; if (num == 0) { res[0] = 0; return res; } res[0] = 0; res[1] = 1; for(int i = 2,j = 2; j < num+1 ;) { for(int k = 0;k < i && j < num+1;j++,k++) res[j] = res[j - i] + 1; i=i*2; } return res; }}
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