338. Counting Bits

Giới thiệu bài toán

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

Yêu cầu

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 <= 105

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++)?

Giải quyết bài toán

C#

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

Python

class Solution(object):
    def countBits(self, n):
        """
        :type n: int
        :rtype: List[int]
        """
        ans = [0] * (n + 1)
        for i in range(1, n + 1):
            ans[i] = ans[i >> 1] + (i & 1)
        return ans

JavaScript

/**
 * @param {number} n
 * @return {number[]}
 */
var countBitsJs = function(n) {
    var ans =  [0];
    for (var i = 1; i<=n; i++) {
        ans[i] = ans[i >> 1] + (i & 1);
    }
    return ans;
};

TypeScript

function countBitsTs(n: number): number[] {
    var ans = [0];
    for (var i = 1; i<=n; i++) {
        ans[i] = ans[i >> 1] + (i & 1);
    }
    return ans;
};

Tham khảo:

• C#
• Python
• JavaScript
• TypeScript