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！