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?
1 /************************************************************************* 2 > File Name: LeetCode338.c 3 > Author: Juntaran 4 > Mail: Jacinthmail@gmail.com 5 > Created Time: 2016年05月10日 星期二 02时33分02秒 6 ************************************************************************/ 7 8 /************************************************************************* 9 10 Given a non negative integer number num. 11 For every numbers i in the range 0 ≤ i ≤ num 12 calculate the number of 1's in their binary representation 13 and return them as an array. 14 15 Example: 16 For num = 5 you should return [0,1,1,2,1,2]. 17 18 Follow up: 19 It is very easy to come up with a solution with run time O(n*sizeof(integer)). 20 But can you do it in linear time O(n) /possibly in a single pass? 21 Space complexity should be O(n). 22 Can you do it like a boss? 23 Do it without using any builtin function like __builtin_popcount in c++ 24 or in any other language. 25 26 Hint: 27 You should make use of what you have produced already. 28 Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. 29 Or does the odd/even status of the number help you in calculating the number of 1s? 30 31 ************************************************************************/ 32 33 #include "stdio.h" 34 35 /** 36 * Return an array of size *returnSize. 37 * Note: The returned array must be malloced, assume caller calls free(). 38 */ 39 40 /* 41 思路： 42 1：2的次幂都是1 43 2：非2的次幂，比如9，找前一个2的次幂比如8，做差，其二进制中1的个数为差+1 44 */ 45 int isPowerOfTwo(int n) 46 { 47 double tmp = log10(n)/log10(2); 48 return tmp == (int)tmp ? 1 : 0; 49 } 50 51 int * countBits(int num, int* returnSize) 52 { 53 int *result = malloc(sizeof(int)*(num+1)); 54 result[0] = 0 ; 55 int last = 0 ; 56 int i; 57 for( i=1 ; i<= num; i++ ) 58 { 59 if(isPowerOfTwo(i)) 60 { 61 result[i] = 1 ; 62 last = i; 63 } 64 else 65 { 66 result[i] = result[i - last] + 1 ; 67 } 68 // printf("%d ",result[i]); 69 } 70 *returnSize = num+1 ; 71 return result; 72 } 73 74 int main() 75 { 76 int* returnSize; 77 countBits(5,returnSize); 78 79 return 0; 80 }