LeetCode-338-bit count

Bit count

Title description: Given a non-negative integer num . For i in the range 0 ≤ i ≤ num , calculate the number of 1s in its binary number and return them as an array.

Please refer to LeetCode official website for example description.

Source: LeetCode
Link: https://leetcode-cn.com/problems/counting-bits/
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Solution one: library function

Being lazy, I directly used the java library function Integer.bitCount solve this problem. Despise yourself

Tip: can use dynamic programming to achieve a more efficient solution.

public class LeetCode_338 {
    /**
     * 使用库函数 Integer.bitCount 直接获取整数对应的二进制的1的个数
     *
     * @param n
     * @return
     */
    public static int[] countBits(int n) {
        int[] result = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            result[i] = Integer.bitCount(i);
        }
        return result;
    }

    public static void main(String[] args) {
        for (int i : countBits(100)) {
            System.out.println(i);
        }
    }
}

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