lc 338 Counting Bits

338 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:

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.

规律公式 Accepted

题干中提示可以利用时间复杂度仅为O(n)的方法来做，说明一定存在着某种规律，不需要把每个数进行除二取模累加这样死做。

经过分析，可以轻松地得知一个数i，它的二进制表示含有1的个数一定等于它除去倒数第一位之外剩余1的个数再加上最后一位是否为1。所以，可易得递推式：ans[i] = ans[i/2] + i%2，为了使得程序运行得更快，我们可以等价地用位运算去替代上式：ans[i] = ans[i >> 1] + (i & 1) 。

class Solution {

public:

    vector<int> countBits(int num) {

        vector<int> ans(num+1, 0);

        for (int i = 1; i <= num; i++)  ans[i] = ans[i >> 1] + (i & 1);

        return ans;

    }

};

巴特西

LN : leetcode 338 Counting Bits

lc 338 Counting Bits

规律公式 Accepted

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