class Solution { public: vector<int> countBits(int num) { vector<int> dp(num+1,0); for(int i=1;i<=num;i<<=1){ for(int j=0;j<i;j++){ if(i+j<=num) dp[i+j] = dp[j]+1; } } return dp; } }; 1 2 3 4 5 6 7 8 9 10 11 12版权...
【LeetCode】338. Counting Bits (2 solutions) Counting Bits 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: Fornum = 5you should return[0,1,1,2,...
https://leetcode.com/problems/counting-bits/ https://leetcode.com/discuss/92796/four-lines-c-time-o-n-space-o-1 https://leetcode.com/discuss/92694/my-408-ms-c-solution-using-bitset https://leetcode.com/discuss/92698/my-448ms-c-easy-solution-o-n-time-and-o-n-space LeetCode All i...
LeetCode "Counting Bits" Neat DP problem to go. frommathimport*classSolution(object):defcountBits(self, num): ret=[0]foriinrange(1, num + 1):ifi & 1:#oddret += [ret[-1] + 1]else: prev= i - 1trailing1s= (i | prev) -i cnt= int(math.log(trailing1s + 1, 2)) ret+= ...
Counting Bits https://leetcode.com/problems/counting-bits/ 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....
代码: class Solution { //338. Counting Bits public int[] countBits(int num Leetcode:338.比特位计数 给定一个非负整数 num。对于 0≤ i≤ num 范围中的每个数字i ,计算其二进制数中的 1 的数目并将它们作为数组返回。 示例 1: 示例 2: 进阶: 给出时间复杂度为O...函数(如 C++ 中的 __...
[leetcode]338. Counting Bits solution 1: 因为乘以2相当于左移,最右添0,所以1的个数都不变。 (这一句的思想是精髓) 奇数i的个数相当于是前一个偶数,末尾+1得到的。而前一个偶数中1的个数=这个偶数/2的1的个数。 又因为偶数和其后一个奇数除以二,都得到同一个偶数。(比如4和2的个数一样,5/2...
ret[i] = ret[i & (i - 1)] + 1. Code classSolution{public:vector<int>countBits(intnum){vector<int>table(num +1,0);for(inti =1; i <= num; i++) { table[i] = table[i & (i -1)] +1; }returntable; } };
ans[i]= ans[i & (i - 1)] + 1;returnans; } } 源码来源:https://leetcode-cn.com/problems/counting-bits/solution/bi-te-wei-ji-shu-by-leetcode/ 这里的x&(x−1)我觉得其实也是用汉明重量的思想。
class Solution { public int[] countBits(int num) { if(num<0) return new int[0]; int[] res=new int[num+1]; Arrays.fill(res, 0); for(int i=1;i<=num;i++) res[i]=res[i&(i-1)]+1; return res; } /* * 这个是使用分解计算 ...