letecode [190] - Reverse Bits

 Reverse bits of a given 32 bits unsigned integer.

 Example 1:

Input: 00000010100101000001111010011100
Output: 00111001011110000010100101000000
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

Input: 11111111111111111111111111111101
Output: 10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.

 

Note:

• Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
• In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above the input represents the signed integer -3 and the output represents the signed integer -1073741825.

题目大意：

 　　求一个32位无符号二进制整数的逆置。

理  解：

 　　方法一：间接法

　　　　从尾部开始遍历二进制整数n。n&1可计算n的最后一位是1还是0；更新n = n>>1。

　　　　用string保存逆置后的结果。每从尾部读取一位即保存在string中，当n=0时，若string长度不满32，则用0补齐。

　　　　将string转换为uint32_t类型即可。

　　方法二：直接法

　　　　二进制整数尾部的数字是逆置后的第一个数字，它逆置后的值为它的值*pow(2,index)，从尾部遍历n，直接计算逆置后的值。

代 码 C++：

方法一：

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        string str = "";
        int len;
        while(n!=0){
            str += to_string(n&1);
            n = n>>1;
        }
        len = str.length();
        while(len<32){
            str += '0';
            len++;
        }
        n = strtoul(str.c_str(),NULL,2);
        return n;
    }
};

方法二：

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t sum=0;
        int index = 31;
        while(n!=0){
            if(n&1==1){
                sum += pow(2,index);    
            }
            index--;
            n = n>>1;
        }
        return sum;
    }
};

运行结果：

 　　方法一：

　　　　执行用时 :4 ms, 在所有C++提交中击败了96.81%的用户

　　　　内存消耗 :8.3 MB, 在所有C++提交中击败了5.12%的用户

　　方法二：

　　　　执行用时 :4 ms, 在所有C++提交中击败了96.81%的用户

　　　　内存消耗 :8.4 MB, 在所有C++提交中击败了5.12%的用户