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
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
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分析
题目要求将一个32位无符号整数对应的二进制串反转，求翻转后的二进制串对应的新的32位无符号整数值。
也就是说，只要我们求得该输入无符号整数的二进制表示形式，将其反转，再转换为十进制即可。
注意的问题，题目明确要求了32位，也就是说对于整数的二进制串必然长度为32，若长度不足则用0补齐。处理后再反转，求新的十进制。
另外，对于此题，我们还需要想到用位运算处理来提高运算速度。
AC代码
class Solution {
public:
//方法一
uint32_t reverseBits(uint32_t n) {
vector<int> bits;
//利用位运算求二进制序列
while (n)
{
bits.push_back(n & 1);
n = n>> 1;
}
//求二进制位反转后对应的十进制数，若bits中长度不足32，以0步之
uint32_t ret = 0;
int size = bits.size();
for (int i = 0 ; i <size; ++i)
{
ret = ret + bits[i] * (1 << (32 - i - 1));
}//for
return ret;
}
//简洁的解法2
uint32_t reverseBits2(uint32_t n) {
uint32_t res = 0;
for (int i = 0; i < 32; ++i) {
res |= (((n >> i) & 1) << (31 - i));
}
return res;
}
};
