LeetCode:Reverse Bits

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;
    }
};