Leetcode 338 Counting Bits
2017-04-29 06:51
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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
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.
撸代码还是需要脑子啊。。
这种是dp的做法。。
港真
popcount 这种做法
真赞
还有一种就是 因为bit数目其实是有规律可循的,在num是2的次方的前面一个达到小高峰,因此可以计算num前面的结果来推算
b其实是从1开始的
P(x+b)=P(x)+1,b=2m>x
return them as an array.
Example:
For
num = 5you 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.
撸代码还是需要脑子啊。。
这种是dp的做法。。
港真
popcount 这种做法
for (count = 0; x != 0; ++count) x &= x - 1; //zeroing out the least significant nonzero bit return count;
真赞
public class Solution { public int[] countBits(int num) { int[] result = new int[num + 1]; for(int i = 1; i <= num; i++){ result[i] = result[i&(i-1)] + 1; } return result; } }
还有一种就是 因为bit数目其实是有规律可循的,在num是2的次方的前面一个达到小高峰,因此可以计算num前面的结果来推算
b其实是从1开始的
P(x+b)=P(x)+1,b=2m>x
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