leecode 解题总结：190. Reverse Bits

#include <iostream>#include <stdio.h>#include <vector>#include <string>#include <bitset>using namespace std;/*问题: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?Related problem: Reverse Integer分析:对给定的无符号32位整数的bit位进行逆置。提供的是无符号整数，如果最后一位为1，那么逆置后1被放到最高位是否要做特殊处理。如果不需要处理，直接用bitset即可，最后转化为unsigned-1=-(1)=-(000000000...1)=111111111...1=2^0 + ... + 2^31=2^32-1=4294967295输入:4326159610-1输出:964176192214748364804294967295关键:1 leecode中bitset不支持at函数，需要用[]*/typedef unsigned uint32_t;class Solution {public:    uint32_t reverseBits(uint32_t n) {        bitset<32> bits(n);//用数n初始化bit数组		int size = bits.size();		for(int i = 0 ; i < 16; i++)		{			int temp = bits[i];			bits[i] = bits[size - i - 1];			bits[size - i - 1] = temp;		}		unsigned result = 0;		for(int i = 0 ; i < size ; i++)		{			result |= (bits[i] << i);		}		return result;    }};void print(vector<int>& result){	if(result.empty())	{		cout << "no result" << endl;		return;	}	int size = result.size();	for(int i = 0 ; i < size ; i++)	{		cout << result.at(i) << " " ;	}	cout << endl;}void process(){	 vector<int> nums;	 uint32_t num;	 Solution solution;	 while(cin >> num )	 {		 uint32_t result = solution.reverseBits(num);		 cout << result << endl;	 }}int main(int argc , char* argv[]){	process();	getchar();	return 0;}