Reverse Bits

190.Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

Note:

  • Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned.
  • In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.

Example 1:

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Input: n = 00000010100101000001111010011100
Output: 964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

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Input: n = 11111111111111111111111111111101
Output: 3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.

位操作:

时间复杂度:O(logn)

空间复杂度:O(1)

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class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t r ans = 0;
for (int i = 0; i < 32 && n > 0; ++i) {
ans |= (n & 1) << (31 - i);
n >>= 1;
}
return ans;
}
};

分治:

时间复杂度:O(1)

空间复杂度:O(1)

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class Solution {
private:
const uint32_t M1 = 0x55555555; // 01010101010101010101010101010101
const uint32_t M2 = 0x33333333; // 00110011001100110011001100110011
const uint32_t M4 = 0x0f0f0f0f; // 00001111000011110000111100001111
const uint32_t M8 = 0x00ff00ff; // 00000000111111110000000011111111

public:
uint32_t reverseBits(uint32_t n) {
n = n >> 1 & M1 | (n & M1) << 1;
n = n >> 2 & M2 | (n & M2) << 2;
n = n >> 4 & M4 | (n & M4) << 4;
n = n >> 8 & M8 | (n & M8) << 8;
return n >> 16 | n << 16;
}
};