LeetCode 190 Reverse Bits (位运算推荐)

最新推荐文章于 2021-03-29 20:17:09 发布

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本文介绍了一种高效的方法来反转一个32位无符号整数的二进制表示，并提供了一个Java代码实现。该方法通过循环将输入整数的每一位依次取出并加入到结果变量中，实现了低位到高位的转换。

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

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.

题目链接：https://leetcode.com/problems/reverse-bits/

题目分析：求解方式有点类似队列的思想，从低位开始取值，先进先出，这样低位就变成了高位

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int ans = 0;
        for (int i = 0; i < 32; i ++) {
            ans <<= 1;
            ans += (n & 1);
            n >>= 1;
        }
        return ans;
    }
}