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:借鉴了java的解答.当num=5时
二进制 1个数
0 --> 0
1 --> 1
10 --> 1
11 --> 2
100 --> 1
101 --> 2
发现规律,后面的数可以利用前面的结果(动态规划,或者叫备忘录).例如数字5,二进制为101,可以将其分为两部分:
(1)最后一位,为1,可以用3%2获得
(2)剩余的数,为10(二进制),可以用5/2=2获得(即右移一位)
则5的bits等于2的bits加上1,即1+1=2.
推广可知,记i的Counting Bits为res[i],则
res[i] = res[i/2] + i%2
代码:
class Solution {
public:
vector<int> countBits(int num) {
vector<int> res(num+1); //初始化输出vector,长度为num+1,int型默认元素为0
for (int i = 0; i <= num; i++) { //一次遍历
res[i] = res[i/2] + i%2;
}
return res;
}
};
