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.
题意&解题思路
输入num,求0-num之间的数中二进制含有1的个数。
题目要求用O(n)的时间复杂度以及O(n)的空间复杂度,即需要线性时间解决问题以及在需要返回的数组空间内直接处理,BF行不通,只能另想办法。
使用 i&(i - 1) 可以将 i 的最后一个数位1消掉,假设我们用 ans[i] 表示数 i 含有多少个1,根据上面的思路我们可以有ans[i] = ans[i & (i - 1)] + 1(消除了最后一个1,所以得另外加回来),则解法如下:
class Solution {
public:
vector<int> countBits(int num) {
vector<int>ans(num + 1, 0);
for(int i = 1; i <= num; i++)ans[i] += ans[i & (i - 1)] + 1;
return ans;
}
};