190. Reverse Bits

190. Reverse Bits

problem

​​190. Reverse Bits​​

solution1:

class Solution {public: uint32_t reverseBits(uint32_t n) { uint32_t res = 0; for(int i=0; i<32; i++) { res = res*2 + n%2; n /=2; } return res; }};

solution2

使用二进制的移位特性实现。

class Solution {public: uint32_t reverseBits(uint32_t n) { n = (n >> 16) | (n << 16); n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8); n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4); n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2); n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1); return n; }};

for 8 bit binary number abcdefgh, the process is as follow:

abcdefgh -> efghabcd -> ghefcdab -> hgfedcba

1. ​​Leetcode_Reverse Bits​​;

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