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.

分析：

這道題完全沒看別人的思路，自己想出來的第一道動態(tài)規(guī)劃題目，開心<(￣︶￣)> 當n為奇數(shù)時，則n-1為偶數(shù)，由二進制可知，n-1的二進制最后一位必定是0 而n的二進制就是在n-1的二進制最后一位加1，并未影響n-1的前面的情況，只把最后一位0,變?yōu)榱?，所以當n=奇數(shù)時，dp[n]=dp[n-1]+1; 舉個栗子：6=1*(2^2)+1*(2^1)+0*（2^0),6的二進制是110.7=1*(2^2)+1*(2^1)+1*（2^0）,7的二進制是111. 當n為偶數(shù)時，先舉個栗子吧，6=1*(2^2)+1*(2^1)+0*（2^0),6的二進制是110. n/2就是把每一項都除以2，但是你會發(fā)現(xiàn)每一項的系數(shù)是不變的，6/2=3=1*(2^1)+1*(2^0)+0(2^0) 所以當n=偶數(shù)時，dp[n]=dp[n/2]. class Solution { public:vector<int> countBits(int num) {vector<int> dp(num+1,0);for(int i=1;i<num+1;i++){if(i%2==0)dp[i]=dp[i/2];else dp[i]=dp[i-1]+1;}return dp;} };

轉(zhuǎn)載于:https://www.cnblogs.com/A-Little-Nut/p/8323535.html

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