题目描述:

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).

翻转一个32位整数的每一个位.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
for(int i = 0; i < 16; i++){
swapIntBit(n, i);
}
return n;
}

void swapIntBit(uint32_t &n, int i){
uint32_t b1 = n & (1 << i);
uint32_t b2 = n & (((unsigned int)0x80000000) >> i);
n &= ~(1 << i);
n &= ~(((unsigned int)0x80000000) >> i);
n |= b1 << (31 - i * 2);
n |= b2 >> (31 - i * 2);
}
};