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:

 1 class Solution {
 2 public:
 3     //2 parts:
 4     //last digit: odd:+1, even:+0
 5     //other digit = Count(remove last digit: i>>1)
 6     vector<int> countBits(int num) {
 7         //int pow2 = 2;
 8         int k=0, j=1;
 9         vector<int> dp(num+1,0);
10         for(int i=1; i<=num; i++) {
11             dp[i]=dp[i>>1];
12             if(i%2) dp[i]+=1;
13         }
14         return dp;
15     }
16 };

 

posted @ 2021-04-07 15:48  habibah_chang  阅读(33)  评论(0编辑  收藏  举报