[LeetCode] Reverse Bits 位操作

weixin_33966095

于 2015-03-09 14:13:00 发布

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原文链接：http://www.cnblogs.com/Azhu/p/4323474.html

本文介绍了如何实现一个高效的32位无符号整数位反转算法，并探讨了针对频繁调用场景的优化策略。通过遍历输入数的位并反向添加，最终得到反转后的整数。同时，提供了代码示例和相关测试用例，旨在提高开发者对于位操作的理解和应用能力。

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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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Bit Manipulation

　　遍历输入数的位，输出数反向添加，循环时候需要注意先移位后修改，反过来会错的。

#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;
}