LeetCode 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:
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.

Hint:

1. You should make use of what you have produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. Or does the odd/even status of the number help you in calculating the number of 1s?

【题目分析】

给定一个数num，返回0 ≤ i ≤ num的所有数的二进制表示形式中包含的‘1’的个数。

【思路】

1. 时间复杂度为O(n*sizeof(integer))的方法

对于0到num中的每个数，如果这个数是奇数，则1的个数加1，然后该数除以2，直到该数为0. 代码如下：

public class Solution {
    public int[] countBits(int num) {
        int[] result = new int[num+1];
        for(int i = 0; i <= num; i++) {
            int count = 0, x = i;
            while(x != 0) {
                if(x % 2 == 1) count++;
                x = x >> 1;
            }
            result[i] = count;
        }
        return result;
    }
}

2. 时间复杂度为O(N)，空间复杂度为O(N)

求解过程中求出的一些结果其实可以直接用在后面求解中，可以大大节省代码时间复杂度，代码如下：

public class Solution {
    public int[] countBits(int num) {
        int[] result = new int[num+1];
        for(int i = 1; i <= num; i++) {
            if(i % 2 == 1) {
                result[i] = result[i>>1] + 1;
            } else {
                result[i] = result[i>>1];
            }
        }
        return result;
    }
}

 3.对上面的代码做进一步的优化

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;
}

java的位运算符要比算术运算符快很多