1、题目
Reverse bits of a given 32 bits unsigned integer.
For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).
Follow up: If this function is called many times, how would you optimize it?
Related problem: Reverse Integer
2、wepon的解法
2.1 解析
这是一道很简单的题目。题意是给定一个整数，要求返回这个整数的二进制形式“反转”之后所得到的整数。看下面的例子，一目了然：
-
整数5的二进制表示为“000...00101”，即2^0+2^2=5
-
整数5的二进制形式反转之后为“101000..000”，得到的整数为2^31+2^29= 2684354560
也就是说，原先指数为0的，反转后指数为31；原先指数为2的，反转后指数为29....以此类推。
2.2 代码
C++
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t result=0;
uint32_t A[32];
for(int i=31;i>=0;i--){
A[31-i] = pow(2.0,i);
}
for(int j=0;j<32;j++){
if(n&1){
result += A[j];
}
n = n>>1;
}
return result;
}
};
python
class Solution:
# @param n, an integer
# @return an integer
def reverseBits(self, n):
table = []
result = 0
for i in range(32)[::-1]:
table.append(2**i)
for j in range(32):
if n&1:
result += table[j]
n = n>>1
return result