LeetCode:Reverse Bits

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Reverse Bits

Total Accepted: 43722 Total Submissions: 150010 Difficulty: Easy

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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思路：

1.假设只有8位，n=0000 1101，要转化成ret=10110000；

2.首先ret=00000000；

3.判断n的最后一位是0还是1，int b=n&1，n>>=1；

4.ret = ret | b，最后一位放到ret的最后一位，ret<<=1；

5.if(i--<8)转3；（这里i初值为8）；

code：

class Solution {public:    uint32_t reverseBits(uint32_t n) {        int ret=0;        int i=32;        while(i--)        {            ret<<=1;// 这里ret先移位，再赋值，因为当n取得最高位的时候，ret不需要再移位            ret |= n&1;            n>>=1;        }        return ret;    }};

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