338. Counting Bits

Given an integer n, return an array ans of length n + 1 such that for each i **(0 <= i <= n) , ans[i] is the number of 1 's in the binary representation of i.

Example 1:

Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10

答案

找规律，奇数的二进制的1个数，与它的一半有关。如3的二进制为11，而1的二进制为01，这相当于1左移1位，然后加一。偶数的二进制的1个数，与它的一半一样，如4的二进制为100，而2的二进制为10，这相当于2左移一位。

go

func countBits(n int) []int {
	res := make([]int,  n+1)
	
	res[0] = 0
	for i := 0; i <= n; i++ {
		if i & 1 == 1{
			res[i] = res[i >> 1] +1
		}else {
			res[i] = res[i >> 1]
		}
	}
	
	return res
}

cpp

class Solution {
public:
    vector<int> countBits(int n) {
        vector<int> res(n+1);
        
        for(int i=1;i<=n;i++){
            if(i&1){
                res[i]=res[i-1]+1;
            } else {
                res[i]=res[i/2];
            }
        }
        return res;
    }
};