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【Leetcode】Counting Bits

时间:2016-05-30 15:34:50      阅读:219      评论:0      收藏:0      [点我收藏+]

题目链接:https://leetcode.com/problems/counting-bits/

题目:

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:
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.
思路:
c[i] :数字i的二进制有多少个1 状态转移方程:
c[i]=c[i>>1]+i&1
比如数字7的二进制111只需要计算最后一位和前两位中1的个数,而前两位的个数和110的个数是一样的,也和11是一样的。

算法:

public int[] countBits(int num) {  
    int res[] = new int[num+1];  
    for(int i=0;i<=num;i++){  
        res[i] = res[i>>1]+(i&1);  
    }  
    return res;  
}  


【Leetcode】Counting Bits

原文:http://blog.csdn.net/yeqiuzs/article/details/51520468

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