原题链接在这里：

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 as 00111001011110000010100101000000).

Follow up:

If this function is called many times, how would you optimize it?

题解：

新学了一种API, Integer.reverse()可以直接返回binary的reverse. 参考链接：

Method 2 res左移一位后加上n的最右一位digit. 

Time Complexity: O(1). 一共左移了32位. Space O(1).

AC Java:

1 public class Solution { 2     // you need treat n as an unsigned value 3     public int reverseBits(int n) { 4         //Method 1 5         //return Integer.reverse(n); 6          7         //Method 2 8         int res = 0; 9         for(int i = 0; i<32; i++){10             res <<= 1;11             res += ((n>>i)&1);12         }13         return res;14     }15 }

Follow up 要求 this function is called many times.

把int 拆成 4个byte 分别reverse 反着加回去. 同时维护HashMap, key是byte, value是reverse 这个byte后的结果int. 

若function 被call了很多次，那么大部分的byte都已经有了对应的reverse value存在HashMap中了。

AC Java:

1 public class Solution { 2     HashMap
     
       hm = new HashMap
      
       (); 3      4     public int reverseBits(int n) { 5         byte [] bytes = new byte[4]; 6          7         //convert int into bytes 8         for(int i = 0 ; i<4; i++){ 9             bytes[i] = (byte)((n>>>8*i) & 0xFF);10         }11         12         int res = 0;13         for(int i = 0; i<4; i++){14             res <<= 8;15             res += reverseByte(bytes[i]);16         }17         18         return res;19     }20     21     private int reverseByte(Byte b){22         if(hm.containsKey(b)){23             return hm.get(b);24         }25         26         int val = 0;27         for(int i = 0; i<8; i++){28             val <<= 1;29             val += ((b>>>i)&1);30         }31         hm.put(b, val);32         return val;33     }34 }
      
     

 类似.