0338. Counting Bits (M)

Counting Bits (M)

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 1:

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

Example 2:

Input: 5
Output: [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.

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题意

计算整数0-num的每一个数的二进制中的1的个数。要求时间复杂度为(O(N))，而非(O(kN))。

思路

个人的做法是针对每一个整数num，它二进制中1的个数，就等于把它最高位的1去掉后得到的整数的二进制中1的个数加1，如8的二进制为1000，而去掉最高位1得到的整数为0，所以8的二进制中1的个数为0+1=1。

更巧妙的方法是对于每个整数i，i&(i-1)的结果就相当于把i最右边的1变为0，这样很容易得到count[i] = count[i&(i-1)] + 1。

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代码实现

Java

去掉最高位

class Solution {
    public int[] countBits(int num) {
        int[] count = new int[num + 1];
        count[0] = 0;
        int size = 1;
        int cnt = 0;
        for (int i = 1; i <= num; i++) {
            if (cnt == size) {
                cnt = 0;
                size *= 2;
            }
            count[i] = count[i - size] + 1;
            cnt++;
        }
        return count;
    }
}

去掉最右1

class Solution {
    public int[] countBits(int num) {
        int[] count = new int[num + 1];
        count[0] = 0;
        for (int i = 1; i <= num; i++) {
            count[i] = count[i & (i - 1)] + 1;
        }
        return count;
    }
}