leetcode：Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000).

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

Related problem: Reverse Integer

分析：题意为反转给定32位无符号整型数的位

思路：我们只需将原整型数从右到左一个一个取出来，然后一个个加到新数的最低位中即可

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

或可参考更简洁做法：

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
for (int i = 0; i < 32; ++i) {
res |= (((n >> i) & 1) << (31 - i));
}
return res;
}
};

其他可参考方法：

#include<iostream>
using namespace std;

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

int main()
{
    uint32_t n = 1;
    Solution sol;
    cout<<sol.reverseBits(n)<<endl;
    return 0;
}

　或：

class Solution {
  public:
    uint32_t reverseBits(uint32_t n) {
    uint32_t m = 0;
    for (int i = 0; i< 32 ; i++,n/=2)
        m = (m<<1) + (n%2);
    return m;
}
 };