【LeetCode】190 Reverse Bits

最新推荐文章于 2022-09-05 21:21:51 发布

原创最新推荐文章于 2022-09-05 21:21:51 发布 · 278 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

LeetCode 专栏收录该内容

29 篇文章

订阅专栏

本文介绍了一种高效实现32位无符号整数位反转的方法，并对比了两种不同的算法实现，指出算法1在性能上优于算法2。同时，提供了对于反复调用此功能时的优化策略。

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?

Note: Although the algorithm 2 solves the problem with one line code (using the original method in Integer.class),the algorithm 1 is faster than 2.

Algorithm 1：（Runtime：2ms）

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int result = 0;
        int index = 1;
        for (int i = 0; i < 32; i++) {
            if ((n & index) == index) {
                result = result | (1 << (31 - i));
            }

            index = index << 1;
        }
        return result;
    }
}

Algorithm 2：（Runtime：3ms）

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        return Integer.reverse(n);
    }
}