【leetcode】338. Counting Bits(C++ & Python)
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338. Counting Bits
题目链接
338.1 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.
338.2 解题思路:
思路一:分为两步,一步计算每个数字的二进制中1的个数,一步将num内所有数字的二进制中1的个数放入vector中。
- 计算二进制中1的个数:根据十进制转二进制定义,如果num大于0,每次都对num取余,如果余数为1,则计数count++。num除以2。遍历结束,返回count。
- 计算num内所有数字的二进制中1的个数:从1到num,依次计算二进制中1的个数并放入vector中。
思路二:(没太想明白,参考leetcode-discuss)初始化vector大小为num+1,且初始值全为0。从1开始遍历vector。每次vector[i]的值,都是由vector中,(i-1)与i按位与后的坐标下的值加1。至于为什么要定位到i-1与i按位与后的坐标,不太清楚。
思路三:初始化vector大小为1,初始值为0。
0—-0个
1—-1个
10—-1个
11—-2个100—-1个
101—-2个
110—-2个
111—-3个从这个序列中我们可以看到规律,在每个二进制开头,1比0多了一个1,10和11分别比0,1多了一个1,100,101,110,111分别比00,01,10,11多了一个1。根据这个规律,以2的指数为界限增加1。所以可设置J=0,循环,j小于num,获取此时vector大小为n,遍历i从0开始,如果碰到i大于n或者j大于num,则停止遍历。否则,vector中push进vector[i]+1。
思路四:初始化vector大小为num+1,且初始值全为0。
0
110
11100
101110
111从序列中可以看出,从1开始,10和11都是在1的后面加了0或1,100和101都是在10的后面加了0或1,110和111都是在11的后面加了0或1,表示第i个位置的个数是由第i/2位置的个数与i本身奇偶性有关。即vector[i]=vector[i/2]+i%2。
思路五:初始化vector大小为num+1,且初始值全为0。思路同思路四,只不过i/2等同于i右移1位,即vector[i]=vector[i>>1]+i%2。
338.3 C++代码:
1、思路一代码(69ms):
class Solution140 {public: int getbit(int num) { int count = 0; while (num>0) { if (num % 2 == 1) count++; num = num / 2; } return count; } vector<int> countBits(int num) { vector<int>bits(num + 1,0); for (int i = 1; i < bits.size();i++) { bits[i] = getbit(i); } return bits; }};
2、思路二代码(83ms)
class Solution140_1 {public: vector<int> countBits(int num) { vector<int>bits(num + 1, 0); for (int i = 1; i < bits.size(); i++) { bits[i] = bits[(i - 1)&i] + 1; } return bits; }};
3、思路三代码(86ms)
class Solution140_2 {public: vector<int> countBits(int num) { vector<int>bits(1, 0); int j = 1; while (j<=num) { int n = bits.size(); for (int i = 0; i < n && j<=num; i++,j++) { bits.push_back(bits[i] + 1); } } return bits; }};
4、思路四代码(93ms)
class Solution140_3 {public: vector<int> countBits(int num) { vector<int>bits(num + 1, 0); for (int i = 1; i < bits.size(); i++) { bits[i] = bits[i / 2] + i % 2; } return bits; }};
5、思路五代码(69ms)
class Solution140_4 {public: vector<int> countBits(int num) { vector<int>bits(num + 1, 0); for (int i = 1; i < bits.size(); i++) { bits[i] = bits[i>>1] + i % 2; } return bits; }};
338.4 Python代码:
1、思路一代码(428ms)
class Solution(object): def countBits(self, num): """ :type num: int :rtype: List[int] """ def getbit(num): count=0 while num>0: if num%2==1: count+=1 num=num/2 return count bits=[0]*(num+1) for i in range(1,num+1): bits[i]=getbit(i) return bits
2、思路二代码(195ms)
class Solution1(object): def countBits(self, num): """ :type num: int :rtype: List[int] """ bits=[0]*(num+1) for i in range(1,num+1): bits[i]=bits[(i-1)&i]+1 return bits
3、思路三代码(199ms)
class Solution2(object): def countBits(self, num): """ :type num: int :rtype: List[int] """ bits=[0] j=1 while j<=num: i=0 n=len(bits) while i<n and j<=num: bits.append(bits[i]+1) i+=1 j+=1 return bits
4、思路四代码(215ms)
class Solution3(object): def countBits(self, num): """ :type num: int :rtype: List[int] """ bits=[0]*(num+1) for i in range(1,num+1): bits[i]=bits[i/2]+i%2 return bits
5、思路五代码(245ms)
class Solution4(object): def countBits(self, num): """ :type num: int :rtype: List[int] """ bits=[0]*(num+1) for i in range(1,num+1): bits[i]=bits[i>>1]+i%2 return bits
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