+(n-1)组的count+1 所以某个数的count就是这个数减所在组2开方数的count+1 res[x]=res[x-pow(2,len(bin(x))-3)]+1 注意python的bin()结果强制带前缀0b 然后类似斐波拉切数列,第一个和第二个强制制定 classSolution(object):defcountBits(self, num): res=[1] * (num+1)ifnum==0: res[0]=...
class Solution(object): def countBits(self, num): """ :type num: int :rtype: List[int] """ s = [0] ans = [0] for i in range(num): c = 0 for i in range(len(s)): if s[i] == 0: s[i] = 1 break else: s[i] = 0 c += 1 if c == len(s): s.append(1)...
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? classSolution(object):defcountBits(self, num): a= [0foriinrange(num+1)]fori...
It is very easy to come up with a solution with run timeO(nsizeof(integer))*. But can you do it in linear timeO(n)/possibly in a single pass? Space complexity should beO(n). Can you do it like a boss? Do it without using any builtin function like__builtin_popcountin c++ or ...
Python3 实现: class Solution: def countBits(self, num: int) -> List[int]: dp = [0] * (num+1) exp = 1 for i in range(1, num+1): if i == 2 ** exp: exp += 1 dp[i] = dp[i-2**(exp-1)] + 1 return dp 方法2: 观察数字规律,总结后可以得出递推公式,如下: f(n)...
class Solution { public: vector<int> countBits(int num) { vector <int> re(num+1, 0); for (int i = 0; i <= num; i++) { re[i] = re[i>>1] + i%2; } return re; } }; 1. 2. 3. 4. 5. 6. 7. 8. 9.
classSolution {public: vector<int> countBits(intnum) { vector<int> ret(num+1,0);for(inti=1;i<=num;i++) ret[i]=ret[i&(i-1)]+1;returnret; } }; 另外,利用上面的思路,当只要我们求正整数n中包含1的个数时,可以这么实现: intcountBit_1(intn) ...
class Solution(object): def countElements(self, arr): """ :type arr: List[int] :rtype: int """ lookup = set(arr) return sum(1 for x in arr if x+1 in lookup) # Time: O(nlogn) # Space: O(1) class Solution(object): def countElements(self, arr): """ :type arr: List[...