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LeetCode-Count 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.
Analysis:
0 | 0+1| 0+1 1+1| 0+1 1+1 1+1 2+1 |..........
0 | 1 | 1 2 | 1 2 2 3 |................
Solution:
public class Solution { public int[] countBits(int num) { int[] res = new int[num + 1]; res[0] = 0; int nextNum = 1; int nextCount = 1; while (nextNum <= num) { for (int i = 0; i < nextCount && nextNum <= num; i++) { res[nextNum++] = res[i] + 1; } nextCount *= 2; } return res; }}
LeetCode-Count Bits
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