请编写一个函数，将一个十进制数字按二进制位进行反转，求出反转位置后的十进制数字。
示例：
输入：417，输出：267。因为417的二进制是110100001，反转后是`100001011。
题目难度：简单
题目来源：Reverse the bits in an integer
def solution(num: int) -> int:
# your code
assert solution(417) == 267
assert solution(267) == 417
assert solution(2017) == 1087
def solution(num: int) -> int:
return int('{0:b}'.format(num)[::-1], 2)
assert solution(417) == 267
assert solution(267) == 417
assert solution(2017) == 1087
def solution_11126(num:int)->int:
‘’’
请编写一个函数，将一个十进制数字按二进制位进行反转，求出反转位置后的十进制数字。
示例：
输入：417，输出：267。因为417的二进制是110100001，反转后是`100001011。
:param num:
:return:
'''
bin_num = bin(num)
bin_str = str(bin_num)
bin_str = bin_str[2:]
reverse_str:str = ""
while bin_str is not "":
reverse_str = reverse_str + bin_str[-1]
bin_str = bin_str[0:-1]
print(reverse_str)
return int(reverse_str, 2)
def test_solution_11126():
assert solution_11126(417)==267
assert solution_11126(1) == 1
assert solution_11126(2) == 1
assert solution_11126(2017) == 1087
def solution(num: int) -> int:
return int((''.join([i for i in bin(num)[2:]][::-1])),2)
assert solution(417) == 267
assert solution(267) == 417
assert solution(2017) == 1087
public int solution(int n) {
ArrayList<Integer> integers = new ArrayList<>();
// 转换为二进制
for (int i = 0; i <= 31; i++) {
integers.add(n >>> i & 1);
}
// 去除多余的位数，并转换为String
String s = "";
for (int i = integers.size() - 1; i >= 0; i--) {
if (integers.get(i) != 0) {
s = integers.stream().limit(i + 1).map(integer -> Integer.toString(integer)).collect(joining(""));
break;
}
}
// 转换为十进制
int b = Integer.parseInt(s);
int decimal = 0;
int p = 0;
while (true) {
if (b == 0) {
break;
} else {
int temp = b % 10;
decimal += temp * Math.pow(2, p);
b = b / 10;
p++;
}
}
return decimal;
}
def solution(num: int):
arr = []
for c in format(num, 'b'):
arr.insert(0, c)
return int(''.join(arr),2)
def solution(num: int) -> int:
# 十进制转二进制后去掉前2位（0b），反转二进制
bin_num = bin(num)[2:][::-1]
# 二进制转十进制
result = int(bin_num, 2)
return result
assert solution(417) == 267
assert solution(267) == 417
assert solution(2017) == 1087