leetcode 338. Counting Bits

ShevaKuilin 發(fā)布于2019-08-16 11:06 / 3008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(shù)字中含有多少個，我們可以使用動態(tài)規(guī)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(shù)或是加上位構(gòu)成的數(shù)字的的個數(shù)。

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(shù)字中含有多少個1，我們可以使用動態(tài)規(guī)劃的方法來計算i中有幾個1。假設(shè)我們已經(jīng)知道前i-1個數(shù)字分別有多少個1，而且i中含有k個數(shù)字，那么其實很容易的想到，i中1的個數(shù)等于前k-1位構(gòu)成的數(shù)字的1的個數(shù)，加上第k位1的個數(shù)，即1或是0。還有一種等價的思路是第0位的1的個數(shù)（0或是1）加上1~k位構(gòu)成的數(shù)字的1的個數(shù)。

    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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