Given a non negative integer number?num. For every numbers?i?in the range?0 ≤ i ≤ num?calculate the number of 1's in their binary representation and return them as an array.

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思路：數字總比[數字二進制最高位置0后的數字]的多一個1

?

public static int[] countBits(int num) {int len = num + 1;int[] res = new int[len];res[0] = 0;for (int i = 1; i < len; i++) {res[i] = res[Util.highBit(i)] + 1;}return res;}//將二進制的最高位置0public static int highBit(int x) {return x - (int) Math.pow(2, (int) (Math.log(x) / Math.log(2)));}

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轉載于:https://www.cnblogs.com/zhaihua/p/5551684.html

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