LeetCode 338. Counting Bits

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题目：

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.

思路：

1.检查每一bit是什么，这是它说很简单的思路

2.偶数的话，是前一个数的1的位数 ＋ 1，偶数的话 / 2直到是奇数，因为两杯之间相对于移位，1的位数相同

代码：C++ 思路2:

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> solve;
        if (num < 0)
            return solve;
        solve.push_back(0);
        if (num == 0){
        return solve;
        }
        
        for(int i = 1;i <= num;i++){
            int index = i;
            if (i % 2){//i是奇数
                solve.push_back(solve.back() + 1);
            }
            else{
                while (index% 2 == 0)
                    index /= 2;
                solve.push_back(solve[index]);
            }
        }
        return solve;
    }
};

LeetCode 338. Counting Bits

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原文地址：http://www.cnblogs.com/gavinxing/p/5392637.html