# LeetCode-Reverse Bits, 1 Bit and number binary correlation

Source: Internet
Author: User

LeetCode-Reverse Bits, 1 Bit and number binary correlation

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as0011100101111000001010010000000 ).

The simple method is to traverse it, but this method is very inefficient and seems to have no better way.

`class Solution {public:    uint32_t reverseBits(uint32_t n) {        if(0 == n)        {            return 0;        }        int res = 0;        for(int i = 0; i < 32; i++)        {            if( n & 1)            {                res += (1 << (31-i));            }            n = n >> 1;        }        return res;    }};`

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 weight ).

For example, the 32-bit integer '11' has binary representation`00000000000000000000000000001011`, So the function shocould return 3.

Note that n & (n-1) can delete the last 1, so you don't have to traverse it all. How many 1 is there and how many times it loops?

`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

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x =-123, return-321 here the calculation process is not complex, but be sure to exceed the range

```Class Solution {public: int reverse (int x) {long xx = x; // prevent negative numbers from exceeding Fan Wen if (0 = x) {return 0 ;} int flag = 0; if (x <0) {flag = 1; xx =-x;} stack

Temp; while (xx) {temp. push (xx % 10); xx = xx/10;} long index = 1; 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 ;}};
```

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