[WXM] LeetCode 338. Counting Bits C++

338. 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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

Approach

1. 题目大意为给你一个区间num，返回0~num中每个数的二进制数中1的数量为多少，这道题很有趣，有两个比较常见解法，一个是借用库bitset可以直接算出1的数量为多少，还有一个解法就是用递推解，dp[0]=0,dp[1]=dp[0]+1,dp[2]=dp[0]+1,dp[3]=dp[1]+1,dp[4]=dp[0]+1,dp[5]=dp[1]+1,dp[6]=dp[2]+1,dp[7]=dp[3]+1,dp[8]=dp[0]+1….

Code

One

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

Two

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int>nums;
        for (int i = 0; i <= num; i++) {
            nums.push_back(bitset<32>(i).count());
        }
        return nums;
    }
};