(*leetcode_位运算)Reverse Bits

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Reverse Bits

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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

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

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注意: x>>1 并没有修改x的值,仍然需要x=x>>1!!!

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


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