原题
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.
Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.
解析
给一个数字，计算从0到该数字的二进制1的个数
不允许使用内置函数
思路
我的解法。。使用了内置函数
最优解：因为*2可以表示为<<2,即一个数的两倍和它的一半的1的个数和这个数相同；也即，若一个数除以2可以除尽，则1的个数就是n/2的1的个数，若除不尽，则为n/2+1
我的解法
public int[] countBits(int num) {
int[] result = new int[num + 1];
for (int i = 0; i <= num; i++) {
result[i] = Integer.bitCount(i);
}
return result;
}
最优解
public int[] countBitsOptimize(int num) {
int[] result = new int[num + 1];
for (int i = 1; i <= num; i++) {
result[i] = result[i >> 1] + (i & 1);
}
return result;
}