338. Counting Bits 仅为学习笔记

标签：lin   cti   number   i++   笔记   single   nbsp   tps   can   

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.

Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

解一：

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int>res(1,0);
        int cnt=0;
        while(cnt<num)
        {
            int sz=res.size();
            for(int i=0;i<sz&&cnt<num;i++,cnt++)
                res.push_back(res[i]+1);
        }
        return res;
    }
};

//思路就是求2,3的时候在前面0,1的基础上加一，再求前4,5,6,7的时候在0,1,2,3的基础上加一，以此类推，注意截止条件

解二：

上面从高位入手找出解决办法，还可以从低位入手。‘1’的个数等于除了最低位之外的‘1’的个数加上最低位‘1’的个数，即ret[n] = ret[n>>1] + n%2，具体代码：

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;  
    }  
};  

338. Counting Bits 仅为学习笔记

标签：lin   cti   number   i++   笔记   single   nbsp   tps   can   

原文地址：http://www.cnblogs.com/wangzao2333/p/7722749.html