publicclassSolution {publicint[] countBits(intnum) {int[] ans =newint[num + 1];for(inti = 1; i <= num; ++i) 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−...
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 i...
ret[i] = ret[i & (i - 1)] + 1. Code class Solution { public: vector<int> countBits(int num) { vector<int> table(num + 1, 0); for (int i = 1; i <= num; i++) { table[i] = table[i & (i - 1)] + 1; } return table; } }; ...
https://leetcode.com/problems/counting-bits/创新互联专注于企业全网营销推广、网站重做改版、天峻网站定制设计、自适应品牌网站建设、H5开发、商城开发、集团公司官网建设、成都外贸网站建设、高端网站制作、响应式网页设计等建站业务,价格优惠性价比高,为天峻等各大城市提供网站开发制作服务。 给定一个非负数n,输出[0...
class Solution{ public: vector<int>countBits(int num){ vector <int> ret(num+1,0); for(int i=0; i<=num;++i){ ret[i]=ret[i>>1]+i%2; } return 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. ...
Credits: Special thanks to@ syedeefor adding this problem and creating all test cases. 代码实现 class Solution: def countBits(self, num): """ :type num: int :rtype: List[int] """ ans=[] for i in range(num+1): ans.append(int(bin(i).count('1'))) ...
Can you solve this real interview question? Counting Bits - Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i. Example 1: Input: n = 2 Output: [
publicint[]countBits(intnum){int[]res=newint[num+1];for(inti=0;i<res.length;i++)res[i]=res[i>>>1]+(i&1);returnres;}
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 __...
File metadata and controls Code Blame 12 lines (11 loc) · 228 Bytes Raw func countBits(n int) []int { dp := make([]int, n + 1) offset := 1 for i := 1; i <= n; i++ { if offset * 2 == i { offset = i } dp[i] = 1 + dp[i - offset] } return dp } 1 ...