190.md

【中规中矩】190. 颠倒二进制位 （位运算）

解题思路

解法1：逐位颠倒。swap(0, 31), swap(1, 30) ... swap(15, 16)

解法2：利用掩码先互换奇偶bit位，再互换奇偶2 bits，再互奇偶4 bits (half byte),再互换奇偶byte，最后互换奇偶16 bits 注意：解法2种python直接翻译的C++，C++能过，而Python却不能，如有Python大神指导原因，麻烦留言告知，先行谢过！

代码

class Solution1 {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t res = 0;
        int INT_LEN = 32;
        for (int i = 0; i < INT_LEN; i++) {
            res |= (n & 0x1) << (INT_LEN - i - 1);
            n >>= 1;
        }      
        return res;
    }
};

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;
    }
};

class Solution1:
    def reverseBits(self, n: int) -> int:
        res = 0
        INT_LEN = 32
        for i  in range(0, INT_LEN) :
            res |= (n & 0x1) << (INT_LEN - i - 1)
            n >>= 1
        
        return res

class Solution:
    def reverseBits(self, n) :
        M1 = 0x55555555 # 01010101010101010101010101010101
        M2 = 0x33333333 # 00110011001100110011001100110011
        M4 = 0x0f0f0f0f # 00001111000011110000111100001111
        M8 = 0x00ff00ff # 00000000111111110000000011111111

        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)

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