[LeetCode338]Counting Bits
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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 num = 5
you should return [0,1,1,2,1,2]
.
对从0到num的每一个数进行模2再除以2的运算,如果对2取模等于1,则1的个数加1,通过这种方法来计算这个数用二进制表示时1的个数,思路比较直接。虽然能够通过测试样例,但时间复杂度相对比较高,为O(nlogn)。希望自己能够找到时间复杂度更低的算法。
class Solution {public: vector<int> countBits(int num) { vector<int> r; r.push_back(0); int count; for(int i=1;i<=num;i++) { count=0; int n=i; while(n>0) { if(n%2==1) { count++; } n/=2; } r.push_back(count); } return r; }};
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