[leetcode] 338. Counting Bits 解题报告

题目链接: 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:
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.

Show Hint 

思路: 主要考察二进制数性质的, 即如果A/2 = B, 那么A比B多了一位, 并且A和B出了A的二进制的右边的一位以外其他都一样, 举个栗子A = 11, 二进制就是1011, B = 5, 二进制是101, 所以我们可以看出其最左边是相等的, 只有A的最后一位不等. 

那么我们可以得出一个结论, 如果A/2  = B, 那么A有多少个1取决于B有多少个1和A最右边一位二进制数是0还是1. 如果A最右边一位是1, 那么A比B多一个1, 否则他们具有相等的1.

代码如下:

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