338.Counting-Bits

338. Counting Bits

้ข˜็›ฎๅœฐๅ€

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

้ข˜็›ฎๆ่ฟฐ

Given a non negative integer number num. For every numbers i in the range 0 โ‰ค i โ‰ค 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.

ไปฃ็ 

Approach #1 DP + Most Significant Bit

ๆ นๆฎ้ข˜็›ฎ็š„่ฆๆฑ‚๏ผŒๆ—ถ้—ดๅ’Œ็ฉบ้—ดๅคๆ‚ๅบฆ๏ผŒๆ˜Žๆ˜พๆ˜ฏ่ฆๅŠจๆ€่ง„ๅˆ’็š„ๆ–นๆณ•

ๅพ—ๅ‡บๆŽจๅˆฐๅ…ฌๅผ๏ผšf(n) = ไธๅคงไบŽf(n)็š„ๆœ€ๅคง็š„2็š„ๆฌกๆ–น +f(k)๏ผŒkไธ€ๅฎšๆ˜ฏๅœจๅ‰้ขๅ‡บ็Žฐ็š„๏ผŒ็”จๆ•ฐ็ป„่ฎฐๅฝ•๏ผŒ็›ดๆŽฅๆŸฅ่ฏข

๏ผŒๆณจๆ„2็š„ๆฌกๆ–น้ƒฝๆ˜ฏไธ€ไธช1๏ผŒ่€Œไธ”ๆ˜ฏๆœ€้ซ˜ไฝ๏ผŒ

  • f(5) = 1 + f(1)

  • f(6) = 1 + f(2)

  • ็›ดๅˆฐ f(8) = 1

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

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)็š„ไฝœ็”จๆ˜ฏๅฐ†ไบŒ่ฟ›ๅˆถ่กจ็คบไธญๅณ่พน็ฌฌไธ€ไธช1็ฝฎ0

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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