LeetCode 第 338 题 (Counting Bits)
2016-05-14 14:39
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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].
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.
借助之前算过的位数来计算,假如ret中保存结果,i的二进制有n位,则ret[i]表示i中bit为1的位数,ret[i >> 1]表示去掉i中最低位的结果,ret[i>>1] + (i&1)即为i的前n-1位为1的数量加上自己最后一位是否为1
class Solution { public: vector<int> countBits(int num) { vector<int> ret; ret.push_back(0); for(int i = 1; i <= num; i++) { int bits = ret[i >> 1] + (i & 1); ret.push_back(bits); } return ret; } };
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