LeetCode: 338. Counting Bits

338. Counting Bits

Description

Given a non-negative integer number num. For every number 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].

题目大概意思是，=：给定一个非负整数num，要求输出从0到num中每个数二进制中1的个数，例如0二进制中1的个数为0，1二进制中1的个数为1，2二进制中1的个数为1，3二进制中1的个数为2。如果num=3，则输出[0,1,1,2]

my program

class Solution {
public:
    int Count(int num)
    {
        int n = 0;
        while (num)
        {
            ++n;
            num = num & (num -1);
        }
        return n;
    }
    vector<int> countBits(int num) {
        vector<int> result;
        for (int i=0; i<=num; ++i)
        {
            result.push_back(Count(i));
        }
        return result;
    }
};

Submission Details
15 / 15 test cases passed.
Status: Accepted
Runtime: 73 ms
You are here!
Your runtime beats 51.45% of cpp submissions.

评价：中规中矩的方法，一个计算输入数二进制中1的个数的函数，从0循环到num将每个数中的1的个数分别压入vector result 中，最后返回结果数组。

other better algorithm

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ret(num+1, 0);
        for (int i = 1; i <= num; ++i)
            ret[i] = ret[i&(i-1)] + 1;
        return ret;
    }
};

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ret(num+1, 0);
        for(int i=1; i<=num; ++i)
            ret[i] = ret[i>>1] + i%2;
        return ret;
    }
};

用到了动态规划。
数组中存储了之前的结果，①当前的i二进制中1的个数等于i&(i-1)二进制中的个数再+1；
或者是②i二进制中1的个数等于i/2二进制的个数再+i%2（即如果i是偶数，那么i二进制中1的个数就等于i/2二进制中1的个数，如果i是奇数，那么i二进制的个数就等于i/2二进制中个数再+1）；