338. Counting Bits

• Total Accepted: 74486 
• Total Submissions: 123045 
• Difficulty: Medium
• Contributor: LeetCode

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.

解题思路：

看完题目之后首先想到的是从每个数字的二进制数着手。对所有n≥1，发现去掉最高位（必然为1）之后，都会和一个小于它的数相同，自然它们的1的个数也相同。那么对应的是哪个小于它的数呢？其实，去掉一个k位二进制数的最高位，就等于这个二进制数减去2^(k-1)。因此，用dp[i]记录i的二进制数中的1的个数，可以得到状态转移方程：

1. dp[0] = 0；

2. dp[i] = dp[i-2^(k-1)] + 1, i > 0.

因此可以得到一个时间复杂度和空间复杂度都是O(n)的算法。

012345678900000001001000110100010101100111100010011122444488

代码：

class Solution {public:    vector<int> countBits(int num) {        vector<int> res(num + 1, 0);        int power = 1;        for(int i = 1; i <= num; i++){            if(power * 2 == i)                power *= 2;            res[i] = res[i - power] + 1;        }        return res;    }};