[Leetcode] Reverse Bits

1Reverse bits of a given, unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represent Ed in binary as 00111001011110000010100101000000).

Follow up:
If This function is called many times, what would you optimize it?

Solving Ideas 1

The procedure is similar to converting a string to shaping, except that it is binary and is processed from the right.

Implementation Code 1

//runtime:10 Ms#include <iostream>#include "inttypes.h"using namespace STD;classSolution { Public: uint32_t reversebits (uint32_t N) {uint32_t result =0;inti = +; while(i--) {result = result *2+ (N &0x1); N >>=1; }returnResult }};intMain () {solution S;cout<<s.reversebits (43261596) <<endl;return 0;}

Solving Ideas 2

uint32_t the length of the machine on the basis of idea 1

Implementation Code 2

//runtime:9 ms  # Include <iostream>   #include "Inttypes.h"  using  namespace  STD ; class  Solution {public : uint32_t reversebits (uint32_t N) {uint32_t result = 0 ; uint32_t i; for  (i = 1 ; I! = 0 ; I <<= 1 ) {result <<= 1 ; Result |= (N & 0x1 ); n >>= 1 ; } return  result; }};

Solving Ideas 3

Solve problems by exchanging bits of unsigned integer numbers directly

Implementation Code 3

//runtime:10 MsclassSolution { Public: uint32_t reversebits (uint32_t N) {uint32_t i; uint32_t x =sizeof(x) *8;//uint32_t number of digits         for(i =0; I < x/2; i++) {swapbit (n, I, X-i-1); }returnN }voidSwapbit (uint32_t& N, uint32_t I, uint32_t j) {uint32_t lo = (n >> i) &0x1; uint32_t hi = (n >> j) &0x1;if(lo ^ hi)//I and J-bits are not equal{//0 ^ 1 = 1            //1 ^ 1 = 0 for Exchange purposesN ^= ((1U<< i) | (1U<< j)); }    }};

[Leetcode] Reverse Bits