LeetCode -- Counting Bits
2016-04-11 10:57
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Question:
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
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.
Analysis:
给定一个非负的整数num,对于0 ≤ i ≤ num的每一个i,计算i的二进制位中有多少个1,返回一个记录1的个数的数组。
Follow up:
你能在O(n)的时间内只遍历一遍得到答案嘛?
空间复杂度为O(n).
思路:
因为是计算1的个数,所以肯定跟除2以及除2后的余数有关系了。而且计算时是递增的,因此下一个数字可以利用上一个数字的计算结果。
代码如下:
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 = 5you 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.
Analysis:
给定一个非负的整数num,对于0 ≤ i ≤ num的每一个i,计算i的二进制位中有多少个1,返回一个记录1的个数的数组。
Follow up:
你能在O(n)的时间内只遍历一遍得到答案嘛?
空间复杂度为O(n).
思路:
因为是计算1的个数,所以肯定跟除2以及除2后的余数有关系了。而且计算时是递增的,因此下一个数字可以利用上一个数字的计算结果。
代码如下:
public class Solution { public int[] countBits(int num) { int[] dp = new int[num+1]; dp[0] = 0; for(int i=1; i<=num; i++) { int t1 = i / 2; int t2 = i % 2; dp[i] = dp[t1] + t2; } return dp; } }
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