Given two integers left and right, return the count of numbers in the inclusive range [left, right] having a prime number of set bits in their binary representation.
Recall that the number of set bits an integer has is the number of 1's present when written in binary.
21 written in binary is 10101, which has 3 set bits.\n
Example 1:
\n\n\nInput: left = 6, right = 10\nOutput: 4\nExplanation:\n6 -> 110 (2 set bits, 2 is prime)\n7 -> 111 (3 set bits, 3 is prime)\n8 -> 1000 (1 set bit, 1 is not prime)\n9 -> 1001 (2 set bits, 2 is prime)\n10 -> 1010 (2 set bits, 2 is prime)\n4 numbers have a prime number of set bits.\n\n\n
Example 2:
\n\n\nInput: left = 10, right = 15\nOutput: 5\nExplanation:\n10 -> 1010 (2 set bits, 2 is prime)\n11 -> 1011 (3 set bits, 3 is prime)\n12 -> 1100 (2 set bits, 2 is prime)\n13 -> 1101 (3 set bits, 3 is prime)\n14 -> 1110 (3 set bits, 3 is prime)\n15 -> 1111 (4 set bits, 4 is not prime)\n5 numbers have a prime number of set bits.\n\n\n
\n
Constraints:
\n\n1 <= left <= right <= 1060 <= right - left <= 104给你两个整数 left 和 right ，在闭区间 [left, right] 范围内，统计并返回 计算置位位数为质数 的整数个数。
计算置位位数 就是二进制表示中 1 的个数。
21 的二进制表示 10101 有 3 个计算置位。\n\n
示例 1：
\n\n\n输入：left = 6, right = 10\n输出：4\n解释：\n6 -> 110 (2 个计算置位，2 是质数)\n7 -> 111 (3 个计算置位，3 是质数)\n9 -> 1001 (2 个计算置位，2 是质数)\n10-> 1010 (2 个计算置位，2 是质数)\n共计 4 个计算置位为质数的数字。\n\n\n
示例 2：
\n\n\n输入：left = 10, right = 15\n输出：5\n解释：\n10 -> 1010 (2 个计算置位, 2 是质数)\n11 -> 1011 (3 个计算置位, 3 是质数)\n12 -> 1100 (2 个计算置位, 2 是质数)\n13 -> 1101 (3 个计算置位, 3 是质数)\n14 -> 1110 (3 个计算置位, 3 是质数)\n15 -> 1111 (4 个计算置位, 4 不是质数)\n共计 5 个计算置位为质数的数字。\n\n\n
\n\n
提示：
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