338. Counting Bits(动态规划）

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：

这道题完全没看别人的思路，自己想出来的第一道动态规划题目，开心<(￣︶￣)>

当n为奇数时，则n-1为偶数，由二进制可知，n-1的二进制最后一位必定是0

而n的二进制就是在n-1的二进制最后一位加1，并未影响n-1的前面的情况，只把最后一位0,变为了1，所以当n=奇数时，dp[n]=dp[n-1]+1;

举个栗子：6=1*(2^2)+1*(2^1)+0*（2^0),6的二进制是110.

         7=1*(2^2)+1*(2^1)+1*（2^0）,7的二进制是111.

当n为偶数时，先举个栗子吧，

         6=1*(2^2)+1*(2^1)+0*（2^0),6的二进制是110.

n/2就是把每一项都除以2，但是你会发现每一项的系数是不变的，

        6/2=3=1*(2^1)+1*(2^0)+0(2^0)

所以当n=偶数时，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;

    }

};

巴特西

338. Counting Bits(动态规划）

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