LeetCode: 338. Counting Bits

友情支持

如果您觉得这个笔记对您有所帮助，看在D瓜哥码这么多字的辛苦上，请友情支持一下，D瓜哥感激不尽，😜

支付宝

有些打赏的朋友希望可以加个好友，欢迎关注D 瓜哥的微信公众号，这样就可以通过公众号的回复直接给我发信息。

wx jikerizhi

公众号的微信号是: jikerizhi。因为众所周知的原因，有时图片加载不出来。如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。

338. Counting Bits

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.

解题分析

从 0 开始，每个奇数包含 1 的个数，都比前面一个数要多 1。所以，只需要计算偶数包含的 1 的个数，然后奇数直接在前面的基础上加 1 即可。

另外一种解法：以 2 的 x 次幂为界，后面的的 x 个数中的 1 的个数，是前面个 x 个数在前面加 1 的结果（位数不够在前面补零，例如从 1 到 5，就是从 1 补零 001，再前面加 1，则为 1001）。

\begin{aligned} (0)&=(0)_{2}\\ (1)&=(1)_{2}\\ (2)&=(10)_{2}\\ (3)&=(11)_{2}\\ P(x+b)&=P(x)+1, b=2^{m}>x\\ \end{aligned}

for (int i = 0; i < 99; i++) {
    int j = 0;
    for (; Math.pow(2, j) <= i; j++) {
        if (Math.pow(2, j) == i && i != 1) {
            System.out.println("---");
        }
    }
    System.out.printf("%2d  %10s\n", i, Integer.toBinaryString(i));
}

用这段代码输出一下，观察一下更明显。

这个题的解法跟 89. Gray Code 有异曲同工之妙！

思考题：尝试"动态规划 + 最低有效位"的解法。

参考资料

比特位计数 - 比特位计数 - 力扣（LeetCode）

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.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
/**
 * Runtime: 1 ms, faster than 99.73% of Java online submissions for Counting Bits.
 * Memory Usage: 43 MB, less than 5.88% of Java online submissions for Counting Bits.
 * <p>
 * Copy from: https://leetcode-cn.com/problems/counting-bits/solution/bi-te-wei-ji-shu-by-leetcode/[比特位计数 - 比特位计数 - 力扣（LeetCode）]
 */
public int[] countBits(int num) {
    int[] result = new int[num + 1];
    int b = 1;
    int i = 0;
    while (b <= num) {
        while (i < b && i + b <= num) {
            result[i + b] = result[i] + 1;
            ++i;
        }
        i = 0;
        b <<= 1;
    }
    return result;
}

/**
 * Runtime: 3 ms, faster than 24.83% of Java online submissions for Counting Bits.
 * Memory Usage: 43 MB, less than 5.88% of Java online submissions for Counting Bits.
 */
public int[] countBitsEvenOdd(int num) {
    int[] result = new int[num + 1];
    for (int i = 0; i <= num; i += 2) {
        result[i] = bitsInNum(i);
    }
    for (int i = 1; i <= num; i += 2) {
        result[i] = result[i - 1] + 1;
    }
    return result;
}

private int bitsInNum(int num) {
    int result = 0;
    while (num != 0) {
        if ((num & 1) == 1) {
            result++;
        }
        num = num >>> 1;
    }
    return result;
}