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 10111111111111111111111111111111.

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.

Follow up:

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

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要求一个无符号整数的二进制表示中’1’的个数。 没必要先把数的二进制表示求出来，只需要将数n和1相与，如果结果为1，说明最低位为1，否则最低位为0，然后不断把n向右移动，直到n为0。 完整代码如下：

class Solution {
public:
    int hammingWeight(uint32_t n)
    {
        int cnt = 0;
        while (n) {
            cnt += n & 1;
            n >>= 1;
        }
        return cnt;
    }
};

本代码提交AC，用时3MS。
二刷。直接n&(n-1)可以去掉n中的最后一个数，所以n有多少个二进制1就运算多少次，性能更佳，代码如下：

public
class Solution { // you need to treat n as an unsigned value
public
    int hammingWeight(int n)
    {
        int ans = 0;
        while (n != 0) {
            ++ans;
            n = n & (n – 1);
        }
        return ans;
    }
}

本代码提交AC，用时2MS。