Leetcode 338. Counting Bits

338. Counting Bits

• Total Accepted: 35688
• Total Submissions: 61823
• Difficulty: Medium

 

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.

 

Hint:

1. You should make use of what you have produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. Or does the odd/even status of the number help you in calculating the number of 1s?

 

 

思路：假设res[i]意味着数字i的二进制中1的个数。

对于数i有res[i*2]=res[i]；res[i*2+1]=res[i]

 

其实也可以写成这样：res[i]=res[i/2] + i%2。

 

代码：

对于数i有，res[i*2]=res[i]，res[i*2+1]=res[i]。

用了一个技巧，O(n/2)的复杂度。

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res(num + 2, 0);//技巧：多申请了一个内存。num为偶数时会用到，num为奇数时用不到。最后都要删除。
        int half = num/2;
        for (int i = 0; i <= half; i++) {
            res[i*2] = res[i];
            res[i*2+1] = res[i] + 1;
        }
        res.pop_back();//不论num是奇数还是偶数，这步都要实施
        return res;
    }
};

 

 

其实也可以写成这样：res[i]=res[i/2] + i%2。

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