# LeetCode Reverse Bits

LeetCode Reverse Bits
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).
If this function is called many times, how would you optimize it?
Related problem: Reverse Integer

```class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t ans = 0;
for (int i = 0; i < 31; i++) {
ans |= (n & 1);
ans <<= 1;
n >>= 1;
}
return ans | (n & 1);
}
};
```

```char tb[16] = {0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15};
uint32_t reverseBits(uint32_t n) {
int curr = 0;
uint32_t ret = 0;
uint32_t msk = 0xF;
for(int i = 0; i < 8; i++) {
ret = ret << 4;
curr = msk&n;
ret |= tb[curr];
n = n >> 4;
}
return ret;
}
```

```      01101001
/         \
0110      1001
/   \     /   \
01   10   10   01
/\   /\   /\   /\
0 1  1 0  1 0  0 1```

```      10010110
/         \
0110      1001
/   \     /   \
10   01   01   10
/\   /\   /\   /\
0 1  1 0  1 0  0 1```

```class Solution {
public:
uint32_t reverseBits(uint32_t x) { // x = ABCDEFGH
x = ((x & 0x55555555) << 1) | ((x & 0xAAAAAAAA) >> 1); // x = B0D0F0H0 | 0A0C0E0G = BADCFEHG
x = ((x & 0x33333333) << 2) | ((x & 0xCCCCCCCC) >> 2); // x = DC00HG00 | 00BA00FE = DCBAHGFE
x = ((x & 0x0F0F0F0F) << 4) | ((x & 0xF0F0F0F0) >> 4); // x = HGFE0000 | 0000DCBA = HGFEDCBA
x = ((x & 0x00FF00FF) << 8) | ((x & 0xFF00FF00) >> 8); // ...
x = ((x & 0x0000FFFF) << 16) | ((x & 0xFFFF0000) >> 16);
return x;
}
};
```

0x55555555 = 0101 0101 0101 0101 0101 0101 0101 0101
0xAAAAAAAA = 1010 1010 1010 1010 1010 1010 1010 1010
0x33333333 = 0011 0011 0011 0011 0011 0011 0011 0011
0xCCCCCCCC = 1100 1100 1100 1100 1100 1100 1100 1100

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