題目：
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.
代碼：
方法一——動態規劃思路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：使用一個mark，記錄下一個將要走到的2的n次方。如果走到了這裏，1的位數變爲0，如果沒有走到這裏，等於1加上數字減去上一個mark時的countBits值。
方法二——動態規劃思路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：當num等於偶數是，其1的個數和該偶數除以2得到的數字的1的個數相同；當num等於奇數時，其1的個數等於該奇數除以2得到的數字的1的個數再加1.要觀察規律。
方法三——動態規劃思路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：我們可以發現每個i值都是 i&(i-1) 對應的值加1。