leetcode 338. Counting Bits

ShevaKuilin 發布于2019-08-16 11:06 / 3003人閱讀

摘要：題目要求思路和代碼這里除了暴力的計算每個數字中含有多少個，我們可以使用動態規劃的方法來計算中有幾個。還有一種等價的思路是第位的的個數或是加上位構成的數字的的個數。

題目要求

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.

思路和代碼

這里除了暴力的計算每個數字中含有多少個1，我們可以使用動態規劃的方法來計算i中有幾個1。假設我們已經知道前i-1個數字分別有多少個1，而且i中含有k個數字，那么其實很容易的想到，i中1的個數等于前k-1位構成的數字的1的個數，加上第k位1的個數，即1或是0。還有一種等價的思路是第0位的1的個數（0或是1）加上1~k位構成的數字的1的個數。

    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;
      }

    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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