LeetCode oj 338. Counting Bits(DP)
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338. Counting Bits QuestionEditorial Solution My Submissions
Total Accepted: 46702
Total Submissions: 79672
Difficulty: Medium
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.
Total Accepted: 46702
Total Submissions: 79672
Difficulty: Medium
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.
输入一个数Num,问从0到Num每个数的二进制表示有多少个1。
第一次做LeetCode上的题,以为这上面的题都是暴力。。。看来想错了= =+,不过也是一道水DP,题目里说有O(n*m)的解法和O(n)的解法,前者很容易想,暴力求1的个数就可以,还可以预处理一下。O(n)的话就是DP了,状态转移方程很好想,dp[i] = dp[i>>1] + i%2;(默默吐槽一句用java写算法果然难受 = =+)
public class Problem338 {public static int[] countBits(int num) {int count [] = new int[num+1]; for(int i=0;i<=num;i++){count[i] = count[i>>1] + i % 2;}return count;}public static void main(String [] args){int re [] = countBits(8);for(int i : re){System.out.println(i);}}}
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