[leetcode] 338. Counting Bits
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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).
- Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.
Hint:
- You should make use of what you have produced already.
- Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
- Or does the odd/even status of the number help you in calculating the number of 1s?
这道题是统计二进制数字num以内每个数中‘1’的个数,题目难度为Medium。
我们先拿num=3为例,结果为[0, 1, 1, 2],得到3以内数字的结果之后,4~7的结果可以在此基础上确定,如下0~3二进制表示为:
bin ret
000 0
001 1
010 1
011 2
4~7二进制表示为:
bin ret
100 1
101 2
110 2
111 3
从上面分析可以看出,4~7除了最高位其他位和0~3是相同的,而最高位是1,这样4~7的结果就可以在0~3的基础上分别加1获得,题目就很容易解决了,具体代码:
class Solution {public: vector<int> countBits(int num) { vector<int> ret(1, 0); int cnt = 0; while(cnt < num) { int sz = ret.size(); for(int i=0; i<sz&&cnt<num; ++i,++cnt) { ret.push_back(ret[i]+1); } } return ret; }};
上面从高位入手找出解决办法,还可以从低位入手。‘1’的个数等于除了最低位之外的‘1’的个数加上最低位‘1’的个数,即ret[n] = ret[n>>1] + n%2,具体代码:
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>>1] + i%2; return ret; }};
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