338. Counting Bits

问题：

求0～num

每个数的二进制中含有 1 的个数。

Example 1:
Input: num = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10

Example 2:
Input: num = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101
 

Constraints:
0 <= num <= 10^5
 

Follow up:

It is very easy to come up with a solution with run time O(32n). Can you do it in linear time O(n) and possibly in a single pass?
Could you solve it in O(n) space complexity?
Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?

　　

解法：DP

1.状态：dp[n]：数字n的二进制有多少个 1。

• 数字 n

2.选择：dp[n]=

• 最后一位：0
  • = 除去最后一位的数：n>>1 = dp[n>>1]
• 最后一位：1
  • = 除去最后一位的数：n>>1 的结果 + 1 = dp[n>>1] + 1

3.base：

• dp[0]=0

代码参考：

class Solution {
public:
    //2 parts:
    //last digit: odd:+1, even:+0
    //other digit = Count(remove last digit: i>>1)
    vector<int> countBits(int num) {
        //int pow2 = 2;
        int k=0, j=1;
        vector<int> dp(num+1,0);
        for(int i=1; i<=num; i++) {
            dp[i]=dp[i>>1];
            if(i%2) dp[i]+=1;
        }
        return dp;
    }
};