leetcode Counting Bits

leetcode 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.

题目地址：leetcode Counting Bits

题意：给定一个数num，求0~num中每个数的1的个数

思路：

方法一

想一想，当一个数为2的整数幂的时候，1的个数为1，比如2（10) 和4(100)，8(1000)

在这之后就是前一个序列的数+1 比如 9(1001) = 1(1) + 8 (1) = 2

就是把一个数分解为小于它的最大2的整数幂 + x

C++

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res = vector<int>(num+1,0);
        int pow2 = 1,before =1;
        for(int i=1;i<=num;i++){
            if (i == pow2){
                before = res[i] = 1;
                pow2 <<= 1;
            }
            else{
                res[i] = res[before] + 1;
                before += 1;
            }
        }
        return res;
    }
};

Java

public class Solution {
    public int[] countBits(int num) {
        int[] res = new int[num+1];
        int pow2 = 1,before =1;
        for(int i=1;i<=num;i++){
            if (i == pow2){
                before = res[i] = 1;
                pow2 <<= 1;
            }
            else{
                res[i] = res[before] + 1;
                before += 1;
            }
        }
        return res;
    }
}

Python

class Solution(object):
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        res = [0] * (num + 1)
        before = pow2 = 1
        for i in xrange(1,num + 1):
            if i == pow2:
                before = res[i] = 1
                pow2 <<= 1
            else:
                res[i] = res[before] + 1
                before += 1
        return res

 

方法二

倒过来想，一个数 * 2 就是把它的二进制全部左移一位，也就是说 1的个数是相等的。

那么我们可以利用这个结论来做。

res[i /2] 然后看看最低位是否为1即可（上面*2一定是偶数，这边比如15和14除以2都是7，但是15时通过7左移一位并且+1得到，14则是直接左移）

所以res[i] = res[i >>1] + (i&1)

C++

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

Java

public class Solution {
    public int[] countBits(int num) {
        int[] res = new int[num+1];
        for(int i=1;i<=num;i++){
            res[i] = res[i >> 1] + (i & 1);
        }
        return res;
    }
}

Python

class Solution(object):
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        res = [0] * (num + 1)
        for i in xrange(1,num + 1):
            res[i] = res[i >> 1] + (i & 1)
        return res

 

本题是leetcode 338  Counting Bits题解，

更多题解可以查看：https://www.hrwhisper.me/leetcode-algorithm-solution/