lc面试准备:Reverse Bits

1 题目

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

接口

public int reverseBits(int n)

uint32_t reverseBits(uint32_t n)

2 思路

简单写一下，Java的思路。Java中是没有无符号整数的，只有有符号的int(0x80000000 ~ 0x7fffffff)。

&和|操作的结合使用。

复杂度

Time: O(n)  Space: O(1)

3 代码

1     public int reverseBits(int n) {

2         int res = 0;

3         for (int i = 0; i < 32; i++) {

4             int bit = (n >> i) & 1;

5             res |= bit << (31 - i);

6         }

7         return res;

8     }

4 总结

(n >> i) & 1是取余数的好方法，不用借助额外的空间。

5 参考

1. leetcode
2. stackoverflow
3. Leetcode: Reverse Bits