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?Credits:Special thanks to @ syedee for adding this problem and creating all test cases.
public class Solution { public int[] countBits(int num) { int[] res = new int[num+1]; int pow = 1; res[0] = 0; for(int i=1,t=0;i<=num;i++,t++) { if(i==pow) { pow = pow * 2; t = 0; } res[i] = res[t] + 1; System.out.println(res[i]); } return res; } }reference: https://discuss.leetcode.com/topic/41785/simple-java-o-n-solution-using-two-pointers/3
转载于:https://www.cnblogs.com/hygeia/p/5706021.html
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