LeetCode OJ-338.Counting Bits

LeetCode OJ-338.Counting Bits

题目描述

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.

题目理解

​ 给定一个数num,求[0, num]之间所有数字的二进制表示各自包含的1的个数,并保存在vector里。可以直接运用位运算进行处理,用1左移0位-31位,去和该数求’&’,’&’是对于左右两边的操作数进行各个位上的逻辑与运算,同1则为1,因此可以用1左移相应位数,做’&’,即可知道每一位是否为1。对[0, num]区间内各数分别做上述操作,即可得到结果集合。

Code

int count_bits_for_one_num(int num)
{
    int cnt = 0;
    int i;
    for (i = 0; i < 32; ++i) {
        if (num & (1 << i)) {
            ++cnt;
        }
    }
    return cnt;
}

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res;
        int i;
        for (i = 0; i <= num; ++i) {
            res.push_back(count_bits_for_one_num(i));
        }
        return res;
    }
};

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