leetcode 338. Counting Bits

ShevaKuilin 发布于2019-08-16 11:06 / 3050人阅读

摘要：题目要求思路和代码这里除了暴力的计算每个数字中含有多少个，我们可以使用动态规划的方法来计算中有几个。还有一种等价的思路是第位的的个数或是加上位构成的数字的的个数。

题目要求

</>复制代码 
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.

思路和代码

这里除了暴力的计算每个数字中含有多少个1，我们可以使用动态规划的方法来计算i中有几个1。假设我们已经知道前i-1个数字分别有多少个1，而且i中含有k个数字，那么其实很容易的想到，i中1的个数等于前k-1位构成的数字的1的个数，加上第k位1的个数，即1或是0。还有一种等价的思路是第0位的1的个数（0或是1）加上1~k位构成的数字的1的个数。

</>复制代码 
    public int[] countBits(int num) {
          int[] ans = new int[num + 1];
          for (int i = 1; i <= num; ++i)
            ans[i] = ans[i & (i - 1)] + 1;
          return ans;
      }

</>复制代码 
    public int[] countBits(int num) {
        int[] res = new int[num+1];
        int cur = 1;
        while(cur <= num){
            res[cur] = 1;
            cur <<= 1;
        }
        
        cur = 1;
        for(int i = 1 ; i<=num ; i++){
            if(res[i] > 0){
                cur = i;
            }else{
                res[i] = res[i-cur] + 1;
            }
        }
        return res;
    }

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