/*
338. Counting 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.
*/
/*
思路：转化成二进制字符串，去掉0，然后统计字符串长度并返回即可。
*/
class Solution {
public int[] countBits(int num) {
int[] nums = new int[num+1];
for(int i=0;i<=num;i++) {
nums[i] = Integer.valueOf( String.valueOf( Integer.toBinaryString( i ) ).replace("0", "").length() );
}
return nums;
}
}
338. Counting 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.
*/
/*
思路：转化成二进制字符串，去掉0，然后统计字符串长度并返回即可。
*/
class Solution {
public int[] countBits(int num) {
int[] nums = new int[num+1];
for(int i=0;i<=num;i++) {
nums[i] = Integer.valueOf( String.valueOf( Integer.toBinaryString( i ) ).replace("0", "").length() );
}
return nums;
}
}
