题目：
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.
代码：
方法一——动态规划思路1：
class Solution {
public:
vector<int> countBits(int num) {
vector<int> res = { 0 };
if (num == 0)return res;
int mark = 1;
for (int i = 1; i <= num; i++) {
if (i == mark) {
res.push_back(1);
mark *= 2;
}
else {
int temp = 1 + res[i - mark/2];
res.push_back(temp);
}
}
return res;
}
};
思路：使用一个mark，记录下一个将要走到的2的n次方。如果走到了这里，1的位数变为0，如果没有走到这里，等于1加上数字减去上一个mark时的countBits值。
方法二——动态规划思路2：
class Solution {
public:
vector<int> countBits(int num) {
vector<int> res{0};
for (int i = 1; i <= num; ++i) {
if (i % 2 == 0) res.push_back(res[i / 2]);
else res.push_back(res[i / 2] + 1);
}
return res;
}
};
思路：当num等于偶数是，其1的个数和该偶数除以2得到的数字的1的个数相同；当num等于奇数时，其1的个数等于该奇数除以2得到的数字的1的个数再加1.要观察规律。
方法三——动态规划思路3
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 & (i - 1)] + 1;
}
return res;
}
};
思路：我们可以发现每个i值都是 i&(i-1) 对应的值加1。