190.reverse-bits.go

/*
 * @lc app=leetcode id=190 lang=golang
 *
 * [190] Reverse Bits
 *
 * https://leetcode.com/problems/reverse-bits/description/
 *
 * algorithms
 * Easy (36.98%)
 * Likes:    1180
 * Dislikes: 421
 * Total Accepted:    287K
 * Total Submissions: 722.2K
 * Testcase Example:  '00000010100101000001111010011100'
 *
 * 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 10111111111111111111111111111111.
 *
 *
 *
 * 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.
 *
 *
 *
 *
 * Follow up:
 *
 * If this function is called many times, how would you optimize it?
 *
 */

// @lc code=start
func reverseBits(num uint32) uint32 {
	return reverseBits1(num)
}

// Bit by Bit, time complexity:O(1), space complexity:O(1)
func reverseBits1(num uint32) uint32 {
	ret, power := uint32(0), uint32(31)
	for num != 0 {
		ret += (num & 1) << power
		num = num >> 1
		power -= 1
	}
	return ret
}

// @lc code=end