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.
给定一个非负整数num。对于范围0≤i≤num中的每个数字i,计算其二进制表示中的1的数目并将它们作为数组返回。
例如 num = 5
应该返回 [0,1,1,2,1,2](
.[0,1,2,3,4,5]的二进制数中‘1’的个数
)
•运行时间为O(n * sizeof(integer))的解决方案非常容易。但是你可以在线性时间O(n)/可能在一次传递中做到吗?
•空间复杂度应该是O(n)。
•在c ++或任何其他语言中不使用__builtin_popcount等内建函数。
class Solution {
public:
vector<int> countBits(int num) {
vector<int> nums(num+1,0);
for(int i = 0;i <= num;i++)
nums[i] = nums[i>>1] + (i&1);
return nums;
}
};
nums[i] = nums[i>>1] + (i&1);
分析:这是一道典型的动态规划问题:nums[i]表示i这个数的二进制表示中的1的数目
以10011001为例。将其分为两部分:
(1)最后一位数(1或0,即“i&1”,相当于“i%2”)
(2)其他数字的二进制表示中的1的数目(1的数字,即“nums[i >> 1]”,相当于“nums [i / 2]”)
注:“+”的运算优先级高于“&”,所以应添加括号
参考:https://leetcode.com/problems/counting-bits/discuss/79539/Three-Line-Java-Solution