标签：
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
基本思路即分别提取出当前最高位和最低位的信息，进行交换即可，然后最高位--，最低位++，直到相遇；32位数，所以0-15和16-31分别进行操作。这边我先判断最高位，如果为1，则判断最低为，如果也为1，不进行操作，如果为0，则最低为变成1，这边用到了位移，然后最高位做减法，则变为0。如果最高位为0，则同理，做相反的操作即可。
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
int cur;
for(int i = 31 ; i >= 16;i--){
if(n&(1<<i)){
cur = n&1<<(31-i);
if(cur == 0){
n -= 1<<i;
n += 1<<(31-i);
}
}
else{
cur = n&1<<(31-i);
if(cur != 0){
n += 1<<i;
n -=1<<(31-i);
}
}
}
return n;
}
};
标签：
原文地址：http://www.cnblogs.com/rockwall/p/5744435.html
