[LeetCode] 190. Reverse Bits 翻转二进制位

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).

Follow up:
If this function is called many times, how would you optimize it?

Related problem: Reverse Integer

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

将输入转换成2进制字符串，再翻转并扩充到32位，再将此32位的二进制转为无符号整数

int 在内存中以二进制形式存储，占据32位。用一个 int 型整数 ans 来记录结果，采用移位操作，因为：1. 注意到移位操作比乘2、除2操作效率更高，2. 移位操作很好地绕开了整型运算的溢出以及符号问题。在每次循环中：ans 每次左移一位，当 n 的最低位为1时，ans 的最低位就变为1，n 每次右移一位。总共进行32次循环。

Java: 每次只需移动1位

class Solution {
    public int reverseBits(int n) {
        int ans = 0;
        for (int i = 0; i < 32; i++) {
            ans <<= 1;
            if ((n & 1) == 1)
                ans++;
            n >>= 1;
        }
        return ans;
    }
}

Java: 第 i 次循环移动 i 位

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

Python:

class Solution:
    # @param n, an integer
    # @return an integer
    def reverseBits(self, n):
        string = bin(n)
        if '-' in string:
            string = string[:3] + string[3:].zfill(32)[::-1]
        else:
            string = string[:2] + string[2:].zfill(32)[::-1]
        return int(string, 2)

Python:

class Solution:
    # @param n, an integer
    # @return an integer
    def reverseBits(self, n):
        result = 0
        for i in xrange(32):
            result <<= 1
            result |= n & 1
            n >>= 1
        return result　　

Python:

def reverseBits(n) :
	res = 0
	while (n > 0) :
		res = res << 1
		if (n & 1 == 1) :
			res = res ^ 1
		n = n >> 1

	return res

Python:

def reverseBits2(n):
    res = 0
    while n > 0:
        res <<= 1
        res |= n & 1
        n >>= 1

    return res 　　

Python: 将n的二进制表示从低位到高位的值依次取出，逆序排列得到翻转后的值。

class Solution(object):
    def reverseBits(self, n):
        """
        :type n: int
        :rtype: int
        """
        res = 0
        for i in xrange(32):
            res <<= 1
            res |= ((n >> i) & 1)
        return res　

Python: 利用Python的bin()函数

class Solution(object):
    def reverseBits(self, n):
        """
        :type n: int
        :rtype: int
        """
        b = bin(n)[:1:-1]
        return int(b + '0'*(32-len(b)), 2)　　

C++:

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t res = 0;
        for (int i = 0; i < 32; ++i) {
            if (n & 1 == 1) {
                res = (res << 1) + 1;
            } else {
                res = res << 1;
            }
            n = n >> 1;
        }
        return res;
    }
};　　

LeetCode中有关位操作的题：

[LeetCode] 191. Number of 1 Bits 二进制数1的个数

Repeated DNA Sequences 求重复的DNA序列

Single Number 单独的数字

Single Number II 单独的数字之二

Grey Code 格雷码

类似题目：

[LeetCode] 191. Number of 1 Bits 二进制数1的个数

[LeetCode] 7. Reverse Integer 翻转整数

All LeetCode Questions List 题目汇总