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Leetcode 190. Reverse Bits

时间:2016-07-30 18:11:25      阅读:217      评论:0      收藏:0      [点我收藏+]

190. Reverse Bits

  • Total Accepted: 71291
  • Total Submissions: 242226
  • Difficulty: Easy

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?

 

思路:二进制数反转。

 

代码:

方法一:

 1 class Solution {
 2 public:
 3     uint32_t reverseBits(uint32_t n) {
 4         n = (n >> 16) | (n << 16);
 5         n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8);
 6         n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4);
 7         n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2);
 8         n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
 9         return n;
10     }
11 };


 

方法二:

 1 class Solution {
 2 public:
 3     uint32_t reverseBits(uint32_t n) {
 4         int i;
 5         uint32_t res=0;
 6         for(i=0;i<32;i++){
 7             res<<=1;
 8             res|=n&1;
 9             n>>=1;
10         }
11         return res;
12     }
13 };

 

Leetcode 190. Reverse Bits

原文:http://www.cnblogs.com/Deribs4/p/5721344.html

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