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.
题目大意
计算0<=x<=num的所有数字，二进制表示里面的1的个数。
解题方法
这个题用DP的方法。
分析规律：
0000 0
-------------
0001 1
-------------
0010 1
0011 2
-------------
0100 1
0101 2
0110 2
0111 3
-------------
1000 1
1001 2
1010 2
1011 3
1100 2
1101 3
1110 3
1111 4
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把第i个数分成两种情况，如果i是偶数那么，它的二进制1的位数等于i/2中1的位数；如果i是奇数，那么，它的二进制1的位数等于i-1的二进制位数+1，又i-1是偶数，所以奇数i的二进制1的位数等于i/2中二进制1的位数+1.
所以上面的这些可以很简单的表达成answer[i] = answer[i >> 1] + (i & 1)。
Python代码如下：
class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
res = [0] * (num + 1)
for i in range(1, num + 1):
res[i] = res[i / 2] + i % 2
return res
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Java代码如下：
public class Solution {
public int[] countBits(int num) {
int[] answer = new int[num+1];
answer[0] = 0;
for(int i = 1; i < answer.length; i++){
answer[i] = answer[i >> 1] + (i & 1);
}
return answer;
}
}
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C++代码如下：
class Solution {
public:
vector<int> countBits(int num) {
vector<int> res(num + 1, 0);
for (int i = 1; i <= num; i ++) {
res[i] = res[i / 2] + i % 2;
}
return res;
}
};
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