leetcode - 338. Counting Bits (bit mannipulation)

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.

my solution:

大致思路:

刚好是2的n次方开始，就可以利用前面的结果了，如。

0000

0001

0010 (开始利用[0,1]的结果，再加上自身最高位1)

0011

0100 (开始利用[0,3]的结果，再加上自身最高位1)

0101

0110

0111

1000  (开始利用[0,7]的结果，再加上自身最高位1)

 ....

class Solution {
public:
	vector<int> countBits(int num) {

		init_arr();

		vector<int> temp(num+1,0);
		//记录当前访问数据最左边1的位置
		int high_bit = 2; 

		temp[0] = 0;
		if (num >=1)
			temp[1] = 1;

		//全部都是基于前面扩展的
		for (int i = 2; i <= num; ++i){
			//arr[2]为4，arr[1]为2
			if (high_bit < 32 && arr[high_bit] <= i)    
				++ high_bit;

			temp[i] = temp[i - arr[high_bit -1] ] + 1;
		}

		return temp;
	}

	static void init_arr(){
		if(0!=init_flag)
			return;

		int value = 1;
		arr[0] = value;

		for (int i = 1; i < ARR_SIZE; ++i)
		{
			value = value << 1;
			arr[i] = value;
		}

		init_flag = 1;
	}

	static const int ARR_SIZE = sizeof(int)*8;
	static int arr[ARR_SIZE];
	static int init_flag;
};

int Solution::init_flag = 0;
int Solution::arr[ARR_SIZE];

C版本的:

 int arr [32] = {1<<0,1<<1,1<<2,1<<3,1<<4,1<<5,1<<6,1<<7,
                 1<<8,1<<9,1<<10,1<<11,1<<12,1<<13,1<<14,1<<15,
                 1<<16,1<<17,1<<18,1<<19,1<<20,1<<21,1<<22,1<<23,
                 1<<24,1<<25,1<<26,1<<27,1<<28,1<<29,1<<30,1<<31};
 
 
int* countBits(int num, int* returnSize) {
    
	//记录当前访问数据最左边1的位置
	int high_bit = 2; 
        int* temp = (int*)malloc(sizeof(int)*(num+1));
	temp[0] = 0;
	*returnSize = num+1;
	if (num >=1)
	    temp[1] = 1;

	//全部都是基于之前的扩展的
	for (int i = 2; i <= num; ++i){
		//arr[2]为4，arr[1]为2
		if (high_bit < 32 && arr[high_bit] <= i)    
			++ high_bit;

		temp[i] = temp[i - arr[high_bit -1] ] + 1;
	}
	return temp;
    
}