LeetCode[338]Counting Bits

mattmu

于 2017-05-15 21:10:42 发布

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文章标签： leetcode dynamic-programming

本文链接：https://blog.youkuaiyun.com/mattmu/article/details/72228929

本文介绍了一种高效算法，用于计算指定范围内每个整数二进制表示中1的个数，并给出C++实现代码。该算法采用动态规划思想，通过复用已计算结果来减少重复计算。

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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).

一开始看此题，感觉无从下手。后来看了几遍，发现还可以，为了减少运算的次数，每次计算要用到上次的结果，这里的话也有些动态规划的思想。

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;
    }
};