Leetcode-reverse Bits, 1 bit and number binary case correlation

https://leetcode.com/problems/reverse-bits/

Reverse bits of a given the unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 ( Represented in binary as00111001011110000010100101000000).

The simple thing is to walk through it, but this method is inefficient, and there seems to be no better way to do it.

Class Solution {public:    uint32_t reversebits (uint32_t N) {        if (0 = = N)        {            return 0;        }        int res = 0;        for (int i = 0; i <; i++)        {            if (N & 1)            {                res + = (1 << (31-i));            }            n = n >> 1;        }        return res;    }};

Number of 1 Bits 

https://leetcode.com/problems/number-of-1-bits/

Write a function that takes an unsigned integer and returns the number of ' 1 ' bits it has (also known as the Hamming weigh T).

For example, the 32-bit integer ' One ' 00000000000000000000000000001011 has a binary representation, so the function should return 3.

Note here that n& (n-1) can eliminate the last 1, so do not have to traverse it all over again, how many 1, how many times

Class Solution {public:    int hammingweight (uint32_t n) {        if (0 = = N)        {            return 0;        }        int res = 0;        while (n)        {            n = n& (n-1);            res++;        }        return res;    }};

Reverse Integer 

https://leetcode.com/problems/reverse-integer/

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return-321 Here the calculation process is not complicated, but notice whether the range is exceeded

Class Solution {public:    int reverse (int x) {        long long xx = x;//Prevent negative numbers from exceeding        if (0 = = x)        {            return 0;        }        int flag = 0;        if (x < 0)        {            flag = 1;            xx =-X;        }        stack<int> temp;        while (xx)        {            temp.push (xx%10);            xx = XX/10;        }        Long long index = 1;        Long long res = 0;        while (!temp.empty ())        {            res + = Temp.top () *index;            index = index*10;            Temp.pop ();        }        if (flag)        {            res = res* ( -1);            if (res < int_min)            return 0;        }        if (res > Int_max)        {            return 0;        }        return res;    }};

Leetcode-reverse Bits, 1 bit and number binary case correlation