题目描述:
给定一个数字n,统计0~n之间的数字二进制的1的个数,并用数组输出
例子:
For num = 5 you should return [0,1,1,2,1,2].
要求:- 算法复杂复o(n)
- 空间复杂度o(n)
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.
思路分析:- 根据题目的要求,时间和空间复杂度,明显是要动态规划的方法
- 得出推到公式:f(n) = 不大于f(n)的最大的2的次方+f(k),k一定是在前面出现的,用数组记录,直接查询
- 举例f(5) = f(4)+ f(1),注意2de次方都是一个1,而且是最高位,f(5) = 1+f(1),f(6) = 1+f(2)直到f(8) = 1
public class Solution {
public int[] countBits(int num) {
int[] res = new int[num+1];
int pow2 = 1,before =1;
for(int i=1;i<=num;i++){
if (i == pow2){
before = res[i] = 1;
pow2 <<= 1;
}
else{
res[i] = res[before] + 1;
before += 1;
}
}
return res;
}
}
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