# 338.Counting-Bits

## ้ข็ฎๅฐๅ

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]

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็ฎ็่ฆๆฑ๏ผๆถ้ดๅ็ฉบ้ดๅคๆๅบฆ๏ผๆๆพๆฏ่ฆๅจๆ่งๅ็ๆนๆณ

ๅพๅบๆจๅฐๅฌๅผ๏ผ`f(n)` = ไธๅคงไบ`f(n)`็ๆๅคง็2็ๆฌกๆน +`f(k)`๏ผkไธๅฎๆฏๅจๅ้ขๅบ็ฐ็๏ผ็จๆฐ็ป่ฎฐๅฝ๏ผ็ดๆฅๆฅ่ฏข

๏ผๆณจๆ2็ๆฌกๆน้ฝๆฏไธไธช1๏ผ่ไธๆฏๆ้ซไฝ๏ผ

• `f(5) = 1 + f(1)`

• `f(6) = 1 + f(2)`

• ็ดๅฐ `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)็ไฝ็จๆฏๅฐไบ่ฟๅถ่กจ็คบไธญๅ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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