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
思路分析:這題比較簡單,基本和Reverse Integer的思路類似,n不斷右移位取末位,結果res不斷左移位和n的末位取或,如此進行32次即可。考察簡單的位運算。
AC Code
public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
//1145
int res = n & 1;
for(int i = 1; i < 32; i++){
n = n>>1;
res = res<<1;
res = res | (n & 1);
}
return res;
}
}