[LeetCode] 338. Counting Bits - 문제풀이

Description

정수가 주어지면 각 ( ) 에 대해 의 이진 표현에서 의 수가 되도록 길이 의 배열n 을 반환 합니다

Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's* in the binary representation of* i.

Example 1:

Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10

Example 2:

Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101

Constraints:

• 0 <= n <= 10^5

Follow up:

• It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
• Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?

Solution 1. Bit Manipulation

public int[] countBits(int n) {
    int[] result = new int[n+1];
    for (int i = 1; i <= n; i++){
        result[i] = result[i >> 1] + (i & 1);
    } 
    return result;
}

Solution 2. bitCount()

public int[] countBits(int n) {
    int[] result = new int[n+1];
    for(int i = 0; i<=n; i++){
        result[i] = (Integer.bitCount(i));
    }
    return result;
}

Reference

Counting Bits - LeetCode