# 338.Counting-Bits

## 338. Counting Bits

## 题目地址

<https://leetcode.com/problems/counting-bits/>

## 题目描述

```
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.
```

## 代码

### Approach #1 DP + Most Significant Bit

根据题目的要求，时间和空间复杂度，明显是要动态规划的方法

得出推到公式：`f(n)` = 不大于`f(n)`的最大的2的次方 +`f(k)`，k一定是在前面出现的，用数组记录，直接查询

，注意2的次方都是一个1，而且是最高位，

* `f(5) = 1 + f(1)`
* `f(6) = 1 + f(2)`
* 直到 `f(8) = 1`

```java
class Solution {
  public int[] countBits(int num) {
        int res = new int[num + 1];
    int pow2 = 1, before = 1;
    for (int i = 1; i <= num; i++) {
      if (i == pow2) {
        res[i] = 1;
        pow2 = pow2 << 1;
        before = 1;
      } else {
        res[i] = res[before] + 1;
        before += 1;
      }
    }

    return res;
  }
}
```

### Approach #2 DP + Least Significant Bit

```java
class Solution {
  public int[] countBits(int num) {
      int[] ans = new int[num + 1];
      for (int i = 1; i <= num; ++i)
        ans[i] = ans[i >> 1] + (i & 1); // x / 2 is x >> 1 and x % 2 is x & 1
      return ans;
  }
}
```

### Approach #3 Pop Count

x & (x - 1)的作用是将二进制表示中右边第一个1置0

```java
class Solution {
  public int[] countBits(int num) {
    int[] res = new int[num + 1];
    for (int i = 0; i <= num; i++) {
      ans[i] = popcount(i);
    }

    return ans;
  }

  private int popcount(int x) {
    int count;
    for (count = 0; x != 0; count++) {
      x &= x - 1; //zeroing out the least significant nonzero bit
    }
    return count;
  }
}
```


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