LeetCode190 Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

Example 1:

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

Example 2:

Input: 11111111111111111111111111111101
Output: 10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.

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 signed integer type and should not affect your implementation, as the internal binary representation of the integer 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.

题源：here；完整实现：here

思路：

我们使用一个uint32_t存储我们的结果，然后利用位移操作反复去原数的最后一位，取32次。

实现1：

	uint32_t reverseBits(uint32_t n) {
		uint32_t res = 0;
		int count = 31;
		while (count >= 0) {
			if (n % 2) {res = res << 1; res += 1;}
			else res = res << 1 + 0;
			n = n>>1;
			count--;
		}
		return res;
	}

实现2：

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	uint32_t reverseBits2(uint32_t n) {
		uint32_t res = 0;
		int count = 31;
		while (count >= 0) {
			res = res << 1;
			res = res | (n & 1);
			n = n >> 1;
			count--;
		}
		return res;
	}