LeetCode-338. Counting Bits

发布时间：2020-12-14 04:23:15 所属栏目：大数据来源：网络整理

导读：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 1: Input: 2Output: [0,1,1] Example 2: Input: 5Output:

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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: 
[0,2,2]

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.

public int[] countBits(int num) {//位运算 my
        int[] res = new int[num+1];
        for (int i = 0; i <=num ; i++) {
            res[i]=0;
            int n = i;
            while (0!=n){
                n= n&(n-1);
                res[i]++;
            }
        }
        return res;
    }

时间复杂度O(n)，找规律

public int[] countBits(int num) {// O(n) my
        int[] res = new int[num+1];
        res[0]=0;
        int flag = 1;
        int j=0;
        if(num>0){
            res[1]=1;
        }
        for (int i = 2; i <=num ; i++) {
            if(i==2*flag){
                flag=flag*2;
                j=0;
            }
            res[i]=res[j]+1;
            j++;
        }
        return res;
    }

简洁版

f[i] = f[i / 2] + i % 2

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

（编辑：李大同）

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