#Leetcode# 338. Counting Bits

https://leetcode.com/problems/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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

代码：

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> dp(num + 1, 0);
        dp[0] = 0;
        for(int i = 1; i <= num; i ++) {
            if(isPowerOfTwo(i)) dp[i] = 1;
            else if(i % 2) dp[i] = dp[i - 1] + 1;
            else dp[i] = dp[i / 2];
        }
        return dp;
    }
    bool isPowerOfTwo(int n) {
        if(n <= 0) return false;
        if((n & (n - 1)) == 0) return true;
        return false;
    }
};

　　一会粗去跑步！

FH 

#Leetcode# 338. Counting Bits

原文：https://www.cnblogs.com/zlrrrr/p/10821575.html